duanlaiyin2356 2014-10-15 02:17
浏览 38

使用ajax帖子时如何从其他人那里获得价值?

How to get value from other when using ajax post ?

i want to get value from input type="text" class="nextpage" in demo.php to input <input type='hidden' id='page_number' name='page' value='1'/> in index.php

i try to do like this code but not work , How can i do that ?

index.php

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<head>
<script src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
</head>





<script>
  $(document).ready(function(){ // onload page for load content //    
        document.getElementById('page_number').value = '1'; 
        $.ajax
        (
            {
                url: 'demo.php',
                type: 'POST',
                data: $('#xxx1').serialize(),
                cache: false,
                success: function (data) {
                    $("#results").append(data); 
                    function_two();
                }
            }
         )
  });
</script>

    <form method="post" id="xxx1" action="Javascript:void(0);">
        <input type='hidden' id='page_number' name='page' value='1'/>
    </form>



<script>
  $(document).ready(function(){ // click button for load content //
      $("#Button").click(function(e){    
        var page = $('#results').find('.nextpage').val();
        document.getElementById('page_number').value = page++; 
        $.ajax
        (
            {
                url: 'demo.php',
                type: 'POST',
                data: $('#xxx2').serialize(),
                cache: false,
                success: function (data) {
                    $("#results").append(data); 
                    function_two();
                }
            }
         )
  });
});
</script>

    <form method="post" id="xxx2" action="Javascript:void(0);">
        <input type='hidden' id='page_number' name='page' value='1'/>
    </form>



<div id="results" style=" width: 100%; "></div>
<input type="submit" id="Button" value="OK">

demo.php

<?PHP
$page = $_POST[page];
$nextpage = ($page+1);
?>
<input type='text' class='nextpage' value='<?PHP echo $nextpage; ?>'>
  • 写回答

2条回答 默认 最新

  • dousongxuan7507 2014-10-15 03:36
    关注

    OK, i get answer , cause of issue from duplicate id name page_number

    just change

    document.getElementById('page_number').value = '1';

    to

    document.getElementById('page_number_1').value = '1';

    and change

    <form method="post" id="xxx1" action="Javascript:void(0);">
        <input type='hidden' id='page_number' name='page' value='1'/>
    </form>
    

    to

    <form method="post" id="xxx1" action="Javascript:void(0);">
        <input type='hidden' id='page_number_1' name='page' value='1'/>
    </form>
    

    it's will work.......

    评论

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