PHP - ajax request to UPDATE mysql database [duplicate]
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This question already has an answer here:
What is the difference between client-side and server-side programming?
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I need help please.
i have 2 files:
- ajax.php
- shows.php
and i have table in my data base that call shows.
in shows table i have column - img_credit.
i try to UPDATE this field.
i insert value to input field and when i click on update button i want the query run and UPDATE my database.
i try:
$(document).ready(function(){
$(".update_credit").click(function(){
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows` SET `img_credits` = '$val_update ' WHERE `id` = show['id'] ");
?>
if i do this i get error but if i change to:
$link->query("UPDATE `shows` SET `img_credits` = 'test' WHERE `id` = 421 ");
it's work but it not help me.
i check and the value shows['img_credit'] is not empty.
thank's
php mysql
marked as duplicate by Akintunde-Rotimi, RiggsFolly
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Nov 8 at 17:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
-1
down vote
favorite
This question already has an answer here:
What is the difference between client-side and server-side programming?
5 answers
I need help please.
i have 2 files:
- ajax.php
- shows.php
and i have table in my data base that call shows.
in shows table i have column - img_credit.
i try to UPDATE this field.
i insert value to input field and when i click on update button i want the query run and UPDATE my database.
i try:
$(document).ready(function(){
$(".update_credit").click(function(){
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows` SET `img_credits` = '$val_update ' WHERE `id` = show['id'] ");
?>
if i do this i get error but if i change to:
$link->query("UPDATE `shows` SET `img_credits` = 'test' WHERE `id` = 421 ");
it's work but it not help me.
i check and the value shows['img_credit'] is not empty.
thank's
php mysql
marked as duplicate by Akintunde-Rotimi, RiggsFolly
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Nov 8 at 17:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
What is the error that you are getting? and where are you setting show['id'] ?
– dsadnick
Nov 8 at 16:59
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
This question already has an answer here:
What is the difference between client-side and server-side programming?
5 answers
I need help please.
i have 2 files:
- ajax.php
- shows.php
and i have table in my data base that call shows.
in shows table i have column - img_credit.
i try to UPDATE this field.
i insert value to input field and when i click on update button i want the query run and UPDATE my database.
i try:
$(document).ready(function(){
$(".update_credit").click(function(){
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows` SET `img_credits` = '$val_update ' WHERE `id` = show['id'] ");
?>
if i do this i get error but if i change to:
$link->query("UPDATE `shows` SET `img_credits` = 'test' WHERE `id` = 421 ");
it's work but it not help me.
i check and the value shows['img_credit'] is not empty.
thank's
php mysql
This question already has an answer here:
What is the difference between client-side and server-side programming?
5 answers
I need help please.
i have 2 files:
- ajax.php
- shows.php
and i have table in my data base that call shows.
in shows table i have column - img_credit.
i try to UPDATE this field.
i insert value to input field and when i click on update button i want the query run and UPDATE my database.
i try:
$(document).ready(function(){
$(".update_credit").click(function(){
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows` SET `img_credits` = '$val_update ' WHERE `id` = show['id'] ");
?>
if i do this i get error but if i change to:
$link->query("UPDATE `shows` SET `img_credits` = 'test' WHERE `id` = 421 ");
it's work but it not help me.
i check and the value shows['img_credit'] is not empty.
thank's
This question already has an answer here:
What is the difference between client-side and server-side programming?
5 answers
php mysql
php mysql
edited Nov 8 at 17:49
RiggsFolly
68.6k1764109
68.6k1764109
asked Nov 8 at 16:52
liran
113
113
marked as duplicate by Akintunde-Rotimi, RiggsFolly
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Nov 8 at 17:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Akintunde-Rotimi, RiggsFolly
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Nov 8 at 17:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
What is the error that you are getting? and where are you setting show['id'] ?
– dsadnick
Nov 8 at 16:59
add a comment |
What is the error that you are getting? and where are you setting show['id'] ?
– dsadnick
Nov 8 at 16:59
What is the error that you are getting? and where are you setting show['id'] ?
– dsadnick
Nov 8 at 16:59
What is the error that you are getting? and where are you setting show['id'] ?
– dsadnick
Nov 8 at 16:59
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
From the code you've provided, it seems that you're mixing up the client side logic (Javascript/jQuery) with the server side logic (PHP).
When you have mixed up client side and server side logic in the same file, the server side code (PHP) will execute first and only after the client side (Javascript). Also in the above code, it doesn't matter if you include PHP code in a Javascript function, the PHP code will execute regardless of where is located in a Javascript structure. That's why is a good practice to avoid mixing them up in the same file.
I would recommend using a basic API structure like this:
let body = "foo"; //param to be sent to server side
let request = $.ajax({
type: "POST",
url: "ajax.php",
data: body
});
request.done(function (response) {
console.log(response);
});
request.fail(function (jqXHR, data) {
console.log("API error", data, jqXHR);
});
And then in your ajax.php file you would have the backend logic to do database operations like this:
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows`
SET `img_credits` = '$val_update'
WHERE `id` = ".$show['id']);
Don't forget to use proper means of protecting against SQL injection, like prepared statements etc.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
From the code you've provided, it seems that you're mixing up the client side logic (Javascript/jQuery) with the server side logic (PHP).
When you have mixed up client side and server side logic in the same file, the server side code (PHP) will execute first and only after the client side (Javascript). Also in the above code, it doesn't matter if you include PHP code in a Javascript function, the PHP code will execute regardless of where is located in a Javascript structure. That's why is a good practice to avoid mixing them up in the same file.
I would recommend using a basic API structure like this:
let body = "foo"; //param to be sent to server side
let request = $.ajax({
type: "POST",
url: "ajax.php",
data: body
});
request.done(function (response) {
console.log(response);
});
request.fail(function (jqXHR, data) {
console.log("API error", data, jqXHR);
});
And then in your ajax.php file you would have the backend logic to do database operations like this:
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows`
SET `img_credits` = '$val_update'
WHERE `id` = ".$show['id']);
Don't forget to use proper means of protecting against SQL injection, like prepared statements etc.
add a comment |
up vote
0
down vote
From the code you've provided, it seems that you're mixing up the client side logic (Javascript/jQuery) with the server side logic (PHP).
When you have mixed up client side and server side logic in the same file, the server side code (PHP) will execute first and only after the client side (Javascript). Also in the above code, it doesn't matter if you include PHP code in a Javascript function, the PHP code will execute regardless of where is located in a Javascript structure. That's why is a good practice to avoid mixing them up in the same file.
I would recommend using a basic API structure like this:
let body = "foo"; //param to be sent to server side
let request = $.ajax({
type: "POST",
url: "ajax.php",
data: body
});
request.done(function (response) {
console.log(response);
});
request.fail(function (jqXHR, data) {
console.log("API error", data, jqXHR);
});
And then in your ajax.php file you would have the backend logic to do database operations like this:
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows`
SET `img_credits` = '$val_update'
WHERE `id` = ".$show['id']);
Don't forget to use proper means of protecting against SQL injection, like prepared statements etc.
add a comment |
up vote
0
down vote
up vote
0
down vote
From the code you've provided, it seems that you're mixing up the client side logic (Javascript/jQuery) with the server side logic (PHP).
When you have mixed up client side and server side logic in the same file, the server side code (PHP) will execute first and only after the client side (Javascript). Also in the above code, it doesn't matter if you include PHP code in a Javascript function, the PHP code will execute regardless of where is located in a Javascript structure. That's why is a good practice to avoid mixing them up in the same file.
I would recommend using a basic API structure like this:
let body = "foo"; //param to be sent to server side
let request = $.ajax({
type: "POST",
url: "ajax.php",
data: body
});
request.done(function (response) {
console.log(response);
});
request.fail(function (jqXHR, data) {
console.log("API error", data, jqXHR);
});
And then in your ajax.php file you would have the backend logic to do database operations like this:
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows`
SET `img_credits` = '$val_update'
WHERE `id` = ".$show['id']);
Don't forget to use proper means of protecting against SQL injection, like prepared statements etc.
From the code you've provided, it seems that you're mixing up the client side logic (Javascript/jQuery) with the server side logic (PHP).
When you have mixed up client side and server side logic in the same file, the server side code (PHP) will execute first and only after the client side (Javascript). Also in the above code, it doesn't matter if you include PHP code in a Javascript function, the PHP code will execute regardless of where is located in a Javascript structure. That's why is a good practice to avoid mixing them up in the same file.
I would recommend using a basic API structure like this:
let body = "foo"; //param to be sent to server side
let request = $.ajax({
type: "POST",
url: "ajax.php",
data: body
});
request.done(function (response) {
console.log(response);
});
request.fail(function (jqXHR, data) {
console.log("API error", data, jqXHR);
});
And then in your ajax.php file you would have the backend logic to do database operations like this:
<?php
$val_update = $show['img_credits'];
$link->query("UPDATE `shows`
SET `img_credits` = '$val_update'
WHERE `id` = ".$show['id']);
Don't forget to use proper means of protecting against SQL injection, like prepared statements etc.
answered Nov 8 at 17:23
Dan D.
1697
1697
add a comment |
add a comment |
What is the error that you are getting? and where are you setting show['id'] ?
– dsadnick
Nov 8 at 16:59