duanhuang2804 2015-07-14 01:34
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Ajax无法正常工作,PHP,AJAX

I'm trying to use ajax on a select tag with 2 options, but it's not getting the $_POST for some reason. It prints out the "---", but it does not print out the $_POST value, which is either 1 or 2. I'm not sure what I did wrong. Please take a look at my code right here, and if you figure out which part of the code is wrong, please point it out with a working example. Thank you.

newtest.php

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>

<script type = "text/javascript"> 

function ajax(url,type,theName,id) {

      $.ajax({
           type: "POST",
           url: url,
           data: { select: $(type+'[name='+theName+']').val()},
           error: function(xhr,status,error){alert(error);},
           success:function(data) {
             document.getElementById( id ).innerHTML = data;
           }

      });

}

</script>

<?php

echo "<select name = 'name' onchange = 'ajax(\"newtestx.php\",\"input\",\"name\",\"output\")'>";
echo "<option value = '1'> 1 </option>";
echo "<option value = '2'> 2 </option>";
echo "</select>";

echo "<div id = 'output'></div>";

?>

newtestx.php

<?php

$name = $_POST['name'];
echo $name."---";

?>
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2条回答 默认 最新

  • dongzaijiao4863 2015-07-14 01:43
    关注

    You need to change this line:

    echo "<select name = 'name' onchange = 'ajax(\"newtestx.php\",\"select\",\"name\",\"output\")'>";
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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