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为什么java脚本不工作?

我有这个代码但我面临铬的错误,没有定义load_new_content_dr ...

<html>
<head>

<script lang="javascript" src="http://code.jquery.com/jquery-latest.js">
  $(document).ready(function(){

    $("#Dr_name").change(load_new_content_dr()); 
    $("#day").change(load_new_content_day());
});
function load_new_content_dr(){
     var selected_option_value=$("#Dr_name option:selected").val();

     $.post("_add_reservasion.php", {option_value: selected_option_value},
         function(data){ 
              $("#day").html(data);
              alert(data);
       }
     );

     $.post("__add_reservasion.php",
      function(data){
        $("#time").html(data);
        alert(data);

      }
    );
}

function load_new_content_day(){
      var selected_day = $("#day option:selected").val();
  $.post("__add_reservasion.php", {selected_day:selected_day},
    function(data){
      $("#time").html(data);
      alert(data);

    }
  );
}

  </script>
</head>
  <body>
 ...

some codes for connecting database and other stiffs...


...
<!--======================doctore name=========================-->
<p>doctor name:</p>
      <select name="Dr_name" id = "Dr_name" form="new" onchange="load_new_content_dr()">
        <option value="null"></option>
        <?php
          while($row = mysqli_fetch_array($table))
          {
            $name = $row["name"]; …
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