2016-10-12 16:39
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I have several dynamic input fields and checkboxes and i would like to disable the input box if the checkbox having a similar id value is checked

This is the php code:

   <table class="table table-striped">

 foreach ($checks as $m => $check) {
    $item ="";
    $checkbox ="";
   $textinput ="";

if ($check->mandatory_customer == 1) { //mandatory customer checks
 $displayx .="<i style='color:red;'>*</i>";
  $item .=  $check->item.$displayx;
   $checkbox .='<input type="checkbox" class="1" id="'.$m.'"' ;
   $textinput .='<input type="text" class="1" id="'.$m.'"' ;

     } else { //not mandatory customer
    $item .=  $check->item;
   $checkbox .='<input type="checkbox" class="0" id="'.$m.'"' ;
 $textinput .='<input type="text" class="0" id="'.$m.'"' ;


echo "<tr id='" . $m . "'>";
echo "<td>" . $m . "</td>";
echo "<td>" . $item . "</td>";
echo "<td>".$checkbox."</td>";
echo "<td>".$textinput."</td>";
echo "</tr>";


Now i would like to disable the input boxes having the same id's as the checkboxes if the checkbox is checked. How do i go through this using jquery

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2条回答 默认 最新

  • dsfds2353 2016-10-12 17:11

    As stated in my comment, it is not permitted to have the same ID on more than one element. You should update the ID on the lines where you assign $checkbox and $textinput so that unique ID values are assigned, which will also help you retrieve the submitted values on the other end. Changing them to something like id="chk_'.$m.'" and id="txt_'.$m.'" would be sufficient.

    Also, it looks like you are not closing your input tags. Add /> to the end of the lines where you assign $checkbox and $textinput, or just > if not using HTML 5.

    Once you've fixed those issues, the following jquery will work for you:

    $(function() {
      $('input[type=checkbox]').change(function() {
        if (this.checked) {
            $(this).closest('tr').find('input[type=text]').prop('disabled', true);
        } else {
            $(this).closest('tr').find('input[type=text]').prop('disabled', false);

    Demo: https://jsfiddle.net/BenjaminRay/nnphq559/

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  • douchun9719 2016-10-12 17:08
        $(document).ready(function () {
            $(":checkbox").on("change", function () {
                if ($(this).prop("checked"))
                    $(":text#"+$(this).attr("id")).prop("disabled", "disabled");
                    $(":text#" + $(this).attr("id")).removeAttr("disabled");
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