Javascript - adding more on.change options

up vote
0
down vote

favorite

I currently have an AJAX script set up on my website which updates a table based on a selection from a dropdown menu [id="afl_player_ID"].

What I want to be able to do though is have the same script run if either [id="afl_player_ID"] or [id="afl_stat_ID"] is changed.

How do I go about adding this functionality into this code?

<script type="text/javascript">

jQuery(document).ready(function ($) {

    var valueCheck;

    jQuery("#afl_player_ID").on("change", function () {

        player_ID = $("#afl_player_ID").val();

        stat_ID = $("#afl_stat_ID").val();

        jQuery.ajax({

            type: "POST",

            url: "/wp-admin/admin-ajax.php",

            data: {

                action: "call_advanced_player_basic_gamelog",

                player_ID: player_ID,

                stat_ID: stat_ID

            },

            success: function (output) {

                jQuery("#advanced_player_basic_gamelog").html(output);

            }

        });

    }).change();

});

</script>

edited Nov 8 at 12:22

ponury-kostek

3,29031121

asked Nov 8 at 12:05

Morts81

1237

1

You can add both IDs and attach change event to both jQuery('#afl_player_ID, #afl_stat_ID').on('change'...
– Farooq Khan
Nov 8 at 12:10

add a comment |

up vote
0
down vote

favorite

I currently have an AJAX script set up on my website which updates a table based on a selection from a dropdown menu [id="afl_player_ID"].

What I want to be able to do though is have the same script run if either [id="afl_player_ID"] or [id="afl_stat_ID"] is changed.

How do I go about adding this functionality into this code?

<script type="text/javascript">

jQuery(document).ready(function ($) {

    var valueCheck;

    jQuery("#afl_player_ID").on("change", function () {

        player_ID = $("#afl_player_ID").val();

        stat_ID = $("#afl_stat_ID").val();

        jQuery.ajax({

            type: "POST",

            url: "/wp-admin/admin-ajax.php",

            data: {

                action: "call_advanced_player_basic_gamelog",

                player_ID: player_ID,

                stat_ID: stat_ID

            },

            success: function (output) {

                jQuery("#advanced_player_basic_gamelog").html(output);

            }

        });

    }).change();

});

</script>

edited Nov 8 at 12:22

ponury-kostek

3,29031121

asked Nov 8 at 12:05

Morts81

1237

1

You can add both IDs and attach change event to both jQuery('#afl_player_ID, #afl_stat_ID').on('change'...
– Farooq Khan
Nov 8 at 12:10

add a comment |

up vote
0
down vote

favorite

I currently have an AJAX script set up on my website which updates a table based on a selection from a dropdown menu [id="afl_player_ID"].

What I want to be able to do though is have the same script run if either [id="afl_player_ID"] or [id="afl_stat_ID"] is changed.

How do I go about adding this functionality into this code?

<script type="text/javascript">

jQuery(document).ready(function ($) {

    var valueCheck;

    jQuery("#afl_player_ID").on("change", function () {

        player_ID = $("#afl_player_ID").val();

        stat_ID = $("#afl_stat_ID").val();

        jQuery.ajax({

            type: "POST",

            url: "/wp-admin/admin-ajax.php",

            data: {

                action: "call_advanced_player_basic_gamelog",

                player_ID: player_ID,

                stat_ID: stat_ID

            },

            success: function (output) {

                jQuery("#advanced_player_basic_gamelog").html(output);

            }

        });

    }).change();

});

</script>

edited Nov 8 at 12:22

ponury-kostek

3,29031121

asked Nov 8 at 12:05

Morts81

1237

I currently have an AJAX script set up on my website which updates a table based on a selection from a dropdown menu [id="afl_player_ID"].

What I want to be able to do though is have the same script run if either [id="afl_player_ID"] or [id="afl_stat_ID"] is changed.

How do I go about adding this functionality into this code?

<script type="text/javascript">

jQuery(document).ready(function ($) {

    var valueCheck;

    jQuery("#afl_player_ID").on("change", function () {

        player_ID = $("#afl_player_ID").val();

        stat_ID = $("#afl_stat_ID").val();

        jQuery.ajax({

            type: "POST",

            url: "/wp-admin/admin-ajax.php",

            data: {

                action: "call_advanced_player_basic_gamelog",

                player_ID: player_ID,

                stat_ID: stat_ID

            },

            success: function (output) {

                jQuery("#advanced_player_basic_gamelog").html(output);

            }

        });

    }).change();

});

</script>

javascript ajax

edited Nov 8 at 12:22

ponury-kostek

3,29031121

asked Nov 8 at 12:05

Morts81

1237

edited Nov 8 at 12:22

ponury-kostek

3,29031121

asked Nov 8 at 12:05

Morts81

1237

edited Nov 8 at 12:22

ponury-kostek

3,29031121

edited Nov 8 at 12:22

ponury-kostek

3,29031121

edited Nov 8 at 12:22

ponury-kostek

3,29031121

asked Nov 8 at 12:05

Morts81

1237

asked Nov 8 at 12:05

Morts81

1237

asked Nov 8 at 12:05

Morts81

1237

1

You can add both IDs and attach change event to both jQuery('#afl_player_ID, #afl_stat_ID').on('change'...
– Farooq Khan
Nov 8 at 12:10

add a comment |

1

You can add both IDs and attach change event to both jQuery('#afl_player_ID, #afl_stat_ID').on('change'...
– Farooq Khan
Nov 8 at 12:10

You can add both IDs and attach change event to both jQuery('#afl_player_ID, #afl_stat_ID').on('change'...
– Farooq Khan
Nov 8 at 12:10

add a comment |

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

Use the code below as suggested by @Farooq Khan in the comments

<script type="text/javascript">

jQuery(document).ready( function($) {

    var valueCheck;

    jQuery('#afl_player_ID,#afl_stat_ID').on( 'change', function () {

         player_ID = $('#afl_player_ID').val();

         stat_ID = $('#afl_stat_ID').val();

     jQuery.ajax({

        type: "POST",

        url: "/wp-admin/admin-ajax.php",

        data: {

            action: 'call_advanced_player_basic_gamelog',

            player_ID: player_ID,

            stat_ID: stat_ID,

        },

         success:function(output){

             jQuery('#advanced_player_basic_gamelog').html( output );

         }

     });

    }).change();

});

</script>

answered Nov 8 at 12:14

Nifal Munzir

26615

Thanks, this works perfectly, I knew it would be something simple. Really appreciate you taking the time to respond.
– Morts81
Nov 9 at 1:32

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

Use the code below as suggested by @Farooq Khan in the comments

<script type="text/javascript">

jQuery(document).ready( function($) {

    var valueCheck;

    jQuery('#afl_player_ID,#afl_stat_ID').on( 'change', function () {

         player_ID = $('#afl_player_ID').val();

         stat_ID = $('#afl_stat_ID').val();

     jQuery.ajax({

        type: "POST",

        url: "/wp-admin/admin-ajax.php",

        data: {

            action: 'call_advanced_player_basic_gamelog',

            player_ID: player_ID,

            stat_ID: stat_ID,

        },

         success:function(output){

             jQuery('#advanced_player_basic_gamelog').html( output );

         }

     });

    }).change();

});

</script>

answered Nov 8 at 12:14

Nifal Munzir

26615

Thanks, this works perfectly, I knew it would be something simple. Really appreciate you taking the time to respond.
– Morts81
Nov 9 at 1:32

add a comment |

up vote
1
down vote

accepted

Use the code below as suggested by @Farooq Khan in the comments

<script type="text/javascript">

jQuery(document).ready( function($) {

    var valueCheck;

    jQuery('#afl_player_ID,#afl_stat_ID').on( 'change', function () {

         player_ID = $('#afl_player_ID').val();

         stat_ID = $('#afl_stat_ID').val();

     jQuery.ajax({

        type: "POST",

        url: "/wp-admin/admin-ajax.php",

        data: {

            action: 'call_advanced_player_basic_gamelog',

            player_ID: player_ID,

            stat_ID: stat_ID,

        },

         success:function(output){

             jQuery('#advanced_player_basic_gamelog').html( output );

         }

     });

    }).change();

});

</script>

answered Nov 8 at 12:14

Nifal Munzir

26615

Thanks, this works perfectly, I knew it would be something simple. Really appreciate you taking the time to respond.
– Morts81
Nov 9 at 1:32

add a comment |

up vote
1
down vote

accepted

Use the code below as suggested by @Farooq Khan in the comments

<script type="text/javascript">

jQuery(document).ready( function($) {

    var valueCheck;

    jQuery('#afl_player_ID,#afl_stat_ID').on( 'change', function () {

         player_ID = $('#afl_player_ID').val();

         stat_ID = $('#afl_stat_ID').val();

     jQuery.ajax({

        type: "POST",

        url: "/wp-admin/admin-ajax.php",

        data: {

            action: 'call_advanced_player_basic_gamelog',

            player_ID: player_ID,

            stat_ID: stat_ID,

        },

         success:function(output){

             jQuery('#advanced_player_basic_gamelog').html( output );

         }

     });

    }).change();

});

</script>

answered Nov 8 at 12:14

Nifal Munzir

26615

Use the code below as suggested by @Farooq Khan in the comments

<script type="text/javascript">

jQuery(document).ready( function($) {

    var valueCheck;

    jQuery('#afl_player_ID,#afl_stat_ID').on( 'change', function () {

         player_ID = $('#afl_player_ID').val();

         stat_ID = $('#afl_stat_ID').val();

     jQuery.ajax({

        type: "POST",

        url: "/wp-admin/admin-ajax.php",

        data: {

            action: 'call_advanced_player_basic_gamelog',

            player_ID: player_ID,

            stat_ID: stat_ID,

        },

         success:function(output){

             jQuery('#advanced_player_basic_gamelog').html( output );

         }

     });

    }).change();

});

</script>

answered Nov 8 at 12:14

Nifal Munzir

26615

answered Nov 8 at 12:14

Nifal Munzir

26615

answered Nov 8 at 12:14

Nifal Munzir

26615

answered Nov 8 at 12:14

Nifal Munzir

26615

Thanks, this works perfectly, I knew it would be something simple. Really appreciate you taking the time to respond.
– Morts81
Nov 9 at 1:32

add a comment |

Thanks, this works perfectly, I knew it would be something simple. Really appreciate you taking the time to respond.
– Morts81
Nov 9 at 1:32

Thanks, this works perfectly, I knew it would be something simple. Really appreciate you taking the time to respond.
– Morts81
Nov 9 at 1:32

add a comment |

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