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Editorial Team
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Asked: 2026-06-11T20:08:54+00:00

I am now using jquery .submit action (form) to POST the data. A normal

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I am now using jquery .submit action (form) to POST the data. A normal POST operation in JAVA does not pass label/value key. For the following (sample) form how I do that ?. I do not want to use Ajax as my form would have upload file fields too and I know how to handle that only in simple POST operation. 

<body>
     <form id="frmRequest" name="frmRequest" >

                <div class="clearfix" id="idRequestDetails"  >
                    <table width="809" border="0" id="tbl_data_1_1_1_1__" summary="Profile">
                      <tr>
                        <th width="156" scope="col"><label class="labelrequest" for="txtProfileName1__">Name</label>
                        </th>
                        <th width="74" scope="col"><label class="labelrequest" for="txtProfileUserID1__">User ID</label></th>
                        <th width="131" scope="col"><label class="labelrequest" for="txtSiteCost1__">Site Cost Centre</label></th>
                         <th width="182" scope="col"><label class="labelrequest" for="txtDetail1__">Additional Details</label></th>
                      </tr>
                      <tr>
                        <td><input type="text" name="txtProfileName1__" id="txtProfileName1__" tabindex="100" /></td>
                        <td><input name="txtProfileUserID1__" type="text" class="clearfix" id="txtProfileUserID1__" tabindex="110" size="8" /></td>
                        <td><input name="txtSiteCost1__" type="text" id="txtSiteCost1__" tabindex="220" size="8" /></td>

                        <td><textarea name="txtDetail1__" rows="1" id="txtDetail1__" tabindex="240"></textarea></td>
                      </tr>
                    </table>
                  </div>
        </body>

Tried the following but not working

foreach ($_POST  as $key => $value)
{
   if($key === 'labels') {
      // Decode JSON string to array
      $value = json_decode($value, true);
   }
   if (!is_array($value))
   {
      $message .= "<br/>".$key." : ".$value;
   }
   else
   {
      foreach ($value as $itemvalue)
      {
         $message .= "<br/>".$value." : ".$itemvalue;
      }
   }
}
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1 Answer

  1. Editorial Team

    You have to build inputs with information you need before form is sent.

    Here is sample where I find text between LABEL tags.
    Create hidden INPUT.
    Set INPUT’s name.
    Set INPUT’s value which is JSON encoded array of label values

    $("form").submit(function(event) {
   $labels = $(this).find("label");
   $ret = [];
   $.each($labels, function() {
       $ret.push($(this).text());
   });
   $input = $("<input>").attr("type", "hidden").attr("name", "labels").val(JSON.stringify($ret));
   $(this).append($input);
});

    Since I put all labels into JSON object, in foreach loop, before first if, you need to find labels data, decode it to array

    Like this:

    foreach ($_POST  as $key => $value)
{
   if($key === 'labels') {
      // Decode JSON string to array
      $value = json_decode($value, true);
   }
   if (!is_array($value))
   {
      $message .= "<br/>".$key." : ".$value;
   }
   else
   {
      foreach ($value as $itemvalue)
      {
         $message .= "<br/>".$value." : ".$itemvalue;
      }
   }
}

    Just in case check out JSON functions and your PHP version

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