CODEWITHSUNDEEP
phphtmlmysqlajaxpost
Danny
Danny

Reputation: 135

Unable to send POST value using AJAX

I have two seperate files, one bargraph.html, and the other data.php.

Bargraph.html is as follows:

<form method="POST" name="dataform" action="">
  <select name="data1" id="data1-value">
    <option value="DateRecorded">DateRecorded</option>
    <option value="InletVoltage">InletVoltage</option>
    <option value="InletCurrent">InletCurrent</option>
    <option value="ActivePower">ActivePower</option>
    <option value="PowerFactor">PowerFactor</option>
    <option value="SystemID">SystemID</option>
    </select>
    <select name="data2" id ="data2-value">
    <option value="DateRecorded">DateRecorded</option>
    <option value="InletVoltage">InletVoltage</option>
    <option value="InletCurrent">InletCurrent</option>
    <option value="ActivePower">ActivePower</option>
    <option value="PowerFactor">PowerFactor</option>
    <option value="SystemID">SystemID</option>
  </select>
</form>

<script type="text/javascript">

$('#data1-value').change(function(){
        var data1Value = $(this).val();
$('#data2-value').change(function(){
        var data2Value = $(this).val();

    $.ajax({
        type: 'post',
        url: 'data.php',
        dataType: 'html',
        data: {data1, data2: data1Value, data2Value},
        success:function(data){ 
            alert(data); 
        }, 
        error:function (xhr, ajaxOptions, thrownError){
            alert(thrownError);
        }, 
        complete: function(){
        }
    });
})});
</script>

A section of data.php is as follows:

if (isset($_POST['data1'])) {
    $opp1 = $_POST['data1'];
    $opp2 = $_POST['data2'];
} else {
    $opp1 = 'SystemID';
    $opp2 = 'ApparentPower';
}

$sql = "SELECT $opp1, $opp2 FROM RaritanMachineDataa";

In my bargraph.html I have two drop down menus. I want the options selected from the drop down menu to be sent using AJAX to my data.php file to perform a select statement on my database.

Currently when I run my code, it's returning an error

Uncaught ReferenceError: data1 is not defined

And this is pointing to line 53:

url: 'data.php',

Could someone give me some assistance on this please as I do not know how to fix this.

UPDATE: (Code below):

<script type="text/javascript">

$('#dataform').submit(function(e){
        var data1Value = $("#data1").val();
        var data2Value = $("#data2").val();

    $.ajax({
        type: 'post',
        url: 'data.php',
        dataType: 'html',
        data: {data1, data1Value, data2: data2Value},
        success:function(data){ 
            alert(data); 
        }, 
        error:function (xhr, ajaxOptions, thrownError){
            alert(thrownError);
        }, 
        complete: function(){
        }
    });
})});
</script>

Returning error:

Uncaught SyntaxError: Unexpected token }

On line (at the 2nd last line)

})});

I have tried removing the bracket, and curly bracket but cannot seem to get it to run. What have I done wrong? Thanks for the assistance

UPDATE (Submit button not POSTING data):

        <option value="PowerFactor">PowerFactor</option>
    <option value="SystemID">SystemID</option>
  </select>
    <input type="submit" name="submit" value="Submit">
</form>

<script type="text/javascript">

    $('#dataform').submit(function(e){
        var data1Value = $("#data1").val();
        var data2Value = $("#data2").val();
    $.ajax({
        type: 'post',
        url: 'data.php',
        dataType: 'html',
        data: {data1, data1Value, data2: data2Value},
        success:function(data){ 
            alert(data); 
        }, 
        error:function (xhr, ajaxOptions, thrownError){
            alert(thrownError);
        }, 
        complete: function(){
        }
    });
    e.preventDefault();
});

Upvotes: 3

Views: 93

Answers (1)

Kebab Programmer
Kebab Programmer

Reputation: 1219

Your Error is here:

  $.ajax({
    type: 'post',
    url: 'data.php',
    dataType: 'html',
    data: {data1, data2: data1Value, data2Value}, <-- Your error is here

Change this line to: 

 data: {data1: data1Value, data2: data2Value}

Also I've just noticed that you are doing this

('#data1-value').change(function(){
    var data1Value = $(this).val();
     $('#data2-value').change(function(){
        var data2Value = $(this).val();

I don't believe that $('#data2-value').change is ever getting called. I suggest you handle the submission of the form after both values have been set!!!

With @Devpro comment, your dropdown menus now need to have an assigned id in order for this submit function to work

HTML

 <select name="data1" id="data1-value"> to  <select id="data1" name="data1" id="data1-value">
 <select name="data2" id ="data2-value"> to  <select id="data2" name="data2" id ="data2-value">

JQuery

 $('#dataform').submit(function(e) {
       var data1Value = $("#data1").val();
       var data2Value = $("#data2").val();

       //ajax submit with my edits
       $.ajax....

       e.preventDefault(); //prevent page refresh.
  });

Upvotes: 3

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