how to uniquely identify tags in react
I am trying to access tags on an article using react. At the backend, I implemented tagging functionality using Django-taggit and the tags are returned as an array of strings.
Then, this is how I have implemented looping through the tags in react
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles">
<li className="d-inline tag-name"
key={tag}
value={tag}>
{tag}
</li>
</Link>
))}
</ul>
The problem is, react gives the error ‘Missing "key" prop for element in iterator’
How do I go about resolving this? It seems I need a unique identifier for each tag for this to work
reactjs django-taggit
add a comment |
I am trying to access tags on an article using react. At the backend, I implemented tagging functionality using Django-taggit and the tags are returned as an array of strings.
Then, this is how I have implemented looping through the tags in react
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles">
<li className="d-inline tag-name"
key={tag}
value={tag}>
{tag}
</li>
</Link>
))}
</ul>
The problem is, react gives the error ‘Missing "key" prop for element in iterator’
How do I go about resolving this? It seems I need a unique identifier for each tag for this to work
reactjs django-taggit
add a comment |
I am trying to access tags on an article using react. At the backend, I implemented tagging functionality using Django-taggit and the tags are returned as an array of strings.
Then, this is how I have implemented looping through the tags in react
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles">
<li className="d-inline tag-name"
key={tag}
value={tag}>
{tag}
</li>
</Link>
))}
</ul>
The problem is, react gives the error ‘Missing "key" prop for element in iterator’
How do I go about resolving this? It seems I need a unique identifier for each tag for this to work
reactjs django-taggit
I am trying to access tags on an article using react. At the backend, I implemented tagging functionality using Django-taggit and the tags are returned as an array of strings.
Then, this is how I have implemented looping through the tags in react
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles">
<li className="d-inline tag-name"
key={tag}
value={tag}>
{tag}
</li>
</Link>
))}
</ul>
The problem is, react gives the error ‘Missing "key" prop for element in iterator’
How do I go about resolving this? It seems I need a unique identifier for each tag for this to work
reactjs django-taggit
reactjs django-taggit
asked Nov 21 '18 at 6:54
frre tyyfrre tyy
5119
5119
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Add a key prop to Link.You don't need key on the li element.You need it at the topmost element of the JSX that your are repeating in map.Keep in mind that the key prop should be a unique constant.If your tags are unique,then just use the tag value or i prefer to do something like this.
Concatenating the tag and index so that even if the tag repeats,the resulting string will be unique.
article.tags.map((tag,index) => (
<Link key = {`${tag}-index`} to="/view-articles">
<li className="d-inline tag-name"
value={tag}>
{tag}
</li>
</Link>
As a sidenote, {${tag}-index
} just means {tag + '' + index}.I prefer ES6 template literals to string concatenation.
disagree on using index there. this way if I put new tag at first position it it would completely recreate the list since nokey
was matched. it's known pitfall of usingindex
as a part ofkey
. Also I believe there will be no duplicated tags since it does not make sense
– skyboyer
Nov 21 '18 at 7:27
I had a scenario where the list value was repeating and i had to render it anyway.So i had to concatenate it.Incase ,you are sure of not repeating then by all means avoid the index.
– anuragb26
Nov 21 '18 at 7:35
then you may use uniqueid
(actually it should be first coming in mind to use withkey
). I can not imagine a case when you can have duplicated values withoutid
and still able to distinguish duplicates when say editing one of them
– skyboyer
Nov 21 '18 at 8:25
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:25
add a comment |
If you look closer at your code, you'll see that the element that is output in the iterator is not the li
element (on which you've placed key
property), but the Link
. Try this:
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles" key={tag}>
<li
className="d-inline tag-name"
value={tag}
>
{tag}
</li>
</Link>
))}
</ul>
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:24
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Add a key prop to Link.You don't need key on the li element.You need it at the topmost element of the JSX that your are repeating in map.Keep in mind that the key prop should be a unique constant.If your tags are unique,then just use the tag value or i prefer to do something like this.
Concatenating the tag and index so that even if the tag repeats,the resulting string will be unique.
article.tags.map((tag,index) => (
<Link key = {`${tag}-index`} to="/view-articles">
<li className="d-inline tag-name"
value={tag}>
{tag}
</li>
</Link>
As a sidenote, {${tag}-index
} just means {tag + '' + index}.I prefer ES6 template literals to string concatenation.
disagree on using index there. this way if I put new tag at first position it it would completely recreate the list since nokey
was matched. it's known pitfall of usingindex
as a part ofkey
. Also I believe there will be no duplicated tags since it does not make sense
– skyboyer
Nov 21 '18 at 7:27
I had a scenario where the list value was repeating and i had to render it anyway.So i had to concatenate it.Incase ,you are sure of not repeating then by all means avoid the index.
– anuragb26
Nov 21 '18 at 7:35
then you may use uniqueid
(actually it should be first coming in mind to use withkey
). I can not imagine a case when you can have duplicated values withoutid
and still able to distinguish duplicates when say editing one of them
– skyboyer
Nov 21 '18 at 8:25
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:25
add a comment |
Add a key prop to Link.You don't need key on the li element.You need it at the topmost element of the JSX that your are repeating in map.Keep in mind that the key prop should be a unique constant.If your tags are unique,then just use the tag value or i prefer to do something like this.
Concatenating the tag and index so that even if the tag repeats,the resulting string will be unique.
article.tags.map((tag,index) => (
<Link key = {`${tag}-index`} to="/view-articles">
<li className="d-inline tag-name"
value={tag}>
{tag}
</li>
</Link>
As a sidenote, {${tag}-index
} just means {tag + '' + index}.I prefer ES6 template literals to string concatenation.
disagree on using index there. this way if I put new tag at first position it it would completely recreate the list since nokey
was matched. it's known pitfall of usingindex
as a part ofkey
. Also I believe there will be no duplicated tags since it does not make sense
– skyboyer
Nov 21 '18 at 7:27
I had a scenario where the list value was repeating and i had to render it anyway.So i had to concatenate it.Incase ,you are sure of not repeating then by all means avoid the index.
– anuragb26
Nov 21 '18 at 7:35
then you may use uniqueid
(actually it should be first coming in mind to use withkey
). I can not imagine a case when you can have duplicated values withoutid
and still able to distinguish duplicates when say editing one of them
– skyboyer
Nov 21 '18 at 8:25
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:25
add a comment |
Add a key prop to Link.You don't need key on the li element.You need it at the topmost element of the JSX that your are repeating in map.Keep in mind that the key prop should be a unique constant.If your tags are unique,then just use the tag value or i prefer to do something like this.
Concatenating the tag and index so that even if the tag repeats,the resulting string will be unique.
article.tags.map((tag,index) => (
<Link key = {`${tag}-index`} to="/view-articles">
<li className="d-inline tag-name"
value={tag}>
{tag}
</li>
</Link>
As a sidenote, {${tag}-index
} just means {tag + '' + index}.I prefer ES6 template literals to string concatenation.
Add a key prop to Link.You don't need key on the li element.You need it at the topmost element of the JSX that your are repeating in map.Keep in mind that the key prop should be a unique constant.If your tags are unique,then just use the tag value or i prefer to do something like this.
Concatenating the tag and index so that even if the tag repeats,the resulting string will be unique.
article.tags.map((tag,index) => (
<Link key = {`${tag}-index`} to="/view-articles">
<li className="d-inline tag-name"
value={tag}>
{tag}
</li>
</Link>
As a sidenote, {${tag}-index
} just means {tag + '' + index}.I prefer ES6 template literals to string concatenation.
answered Nov 21 '18 at 7:22
anuragb26anuragb26
36926
36926
disagree on using index there. this way if I put new tag at first position it it would completely recreate the list since nokey
was matched. it's known pitfall of usingindex
as a part ofkey
. Also I believe there will be no duplicated tags since it does not make sense
– skyboyer
Nov 21 '18 at 7:27
I had a scenario where the list value was repeating and i had to render it anyway.So i had to concatenate it.Incase ,you are sure of not repeating then by all means avoid the index.
– anuragb26
Nov 21 '18 at 7:35
then you may use uniqueid
(actually it should be first coming in mind to use withkey
). I can not imagine a case when you can have duplicated values withoutid
and still able to distinguish duplicates when say editing one of them
– skyboyer
Nov 21 '18 at 8:25
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:25
add a comment |
disagree on using index there. this way if I put new tag at first position it it would completely recreate the list since nokey
was matched. it's known pitfall of usingindex
as a part ofkey
. Also I believe there will be no duplicated tags since it does not make sense
– skyboyer
Nov 21 '18 at 7:27
I had a scenario where the list value was repeating and i had to render it anyway.So i had to concatenate it.Incase ,you are sure of not repeating then by all means avoid the index.
– anuragb26
Nov 21 '18 at 7:35
then you may use uniqueid
(actually it should be first coming in mind to use withkey
). I can not imagine a case when you can have duplicated values withoutid
and still able to distinguish duplicates when say editing one of them
– skyboyer
Nov 21 '18 at 8:25
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:25
disagree on using index there. this way if I put new tag at first position it it would completely recreate the list since no
key
was matched. it's known pitfall of using index
as a part of key
. Also I believe there will be no duplicated tags since it does not make sense– skyboyer
Nov 21 '18 at 7:27
disagree on using index there. this way if I put new tag at first position it it would completely recreate the list since no
key
was matched. it's known pitfall of using index
as a part of key
. Also I believe there will be no duplicated tags since it does not make sense– skyboyer
Nov 21 '18 at 7:27
I had a scenario where the list value was repeating and i had to render it anyway.So i had to concatenate it.Incase ,you are sure of not repeating then by all means avoid the index.
– anuragb26
Nov 21 '18 at 7:35
I had a scenario where the list value was repeating and i had to render it anyway.So i had to concatenate it.Incase ,you are sure of not repeating then by all means avoid the index.
– anuragb26
Nov 21 '18 at 7:35
then you may use unique
id
(actually it should be first coming in mind to use with key
). I can not imagine a case when you can have duplicated values without id
and still able to distinguish duplicates when say editing one of them– skyboyer
Nov 21 '18 at 8:25
then you may use unique
id
(actually it should be first coming in mind to use with key
). I can not imagine a case when you can have duplicated values without id
and still able to distinguish duplicates when say editing one of them– skyboyer
Nov 21 '18 at 8:25
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:25
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:25
add a comment |
If you look closer at your code, you'll see that the element that is output in the iterator is not the li
element (on which you've placed key
property), but the Link
. Try this:
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles" key={tag}>
<li
className="d-inline tag-name"
value={tag}
>
{tag}
</li>
</Link>
))}
</ul>
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:24
add a comment |
If you look closer at your code, you'll see that the element that is output in the iterator is not the li
element (on which you've placed key
property), but the Link
. Try this:
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles" key={tag}>
<li
className="d-inline tag-name"
value={tag}
>
{tag}
</li>
</Link>
))}
</ul>
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:24
add a comment |
If you look closer at your code, you'll see that the element that is output in the iterator is not the li
element (on which you've placed key
property), but the Link
. Try this:
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles" key={tag}>
<li
className="d-inline tag-name"
value={tag}
>
{tag}
</li>
</Link>
))}
</ul>
If you look closer at your code, you'll see that the element that is output in the iterator is not the li
element (on which you've placed key
property), but the Link
. Try this:
<ul className="tags" >
{!article.tags ? "" : article.tags.map(tag => (
<Link to="/view-articles" key={tag}>
<li
className="d-inline tag-name"
value={tag}
>
{tag}
</li>
</Link>
))}
</ul>
answered Nov 21 '18 at 7:25
CerberusCerberus
1,050820
1,050820
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:24
add a comment |
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:24
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:24
Thanks, this resolved the issue.
– frre tyy
Nov 21 '18 at 11:24
add a comment |
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