duanditang2916 2017-06-03 03:43
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在ajax php中没有提交表单的唯一ID

I have a form there are 5 fields where I want to automatic generate roll number in 5th field after filling up 4 fields(Name,Phone,Course,Batch) without submitting form.but after filling up 4 field no value comes in 5th field(roll).Below is my code

index.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" <head>
<title>Untitled Document</title>
<script>
    function my_validate_func() {
        if ($('#name').val() != "" && $('#phone').val() != "" &&
            $('#course').val() != "" && $('#center').val() != "") {
            $.ajax({
                type: "POST",
                url: 'submit.php',
                success: function(response) {
                    $('#roll').val(response.roll);
                }
            });
        }
    }
</script>
</head>

<body>
    <form method="post" action="">
        <input type="text" name="name" id="name" onchange="my_validate_func()">
        <input type="text" name="phone" id="phone" onchange="my_validate_func()">
        <input type="text" name="course" id="course" onchange="my_validate_func()">
        <input type="text" name="center" id="center" onchange="my_validate_func()">
        <input type="text" name="roll" id="roll" value="<?php $roll; ?>">
    </form>
</body>

</html>

**submit.php code is below**

<?php
$roll=rand(100000,999999);
echo $roll;
?>
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2条回答 默认 最新

  • douningqiu4991 2017-06-03 05:22
    关注
    Corrected Code,
    
    <!DOCTYPE> 
    <head>
    <title>Untitled Document</title>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
    <script type='text/javascript'>
    $( document ).ready(function() {});
        function my_validate_func() {
            if ($('#name').val() != "" && $('#phone').val() != "" &&
                $('#course').val() != "" && $('#center').val() != "") {
                $.ajax({
                    type: "POST",
                    url: 'submit.php',
                    success: function(response) {
                        $('#roll').val(response);
                    }
                });
            }
        }
    </script>
    </head>
    
    <body>
        <form method="post" action="">
            <input type="text" name="name" id="name" onchange="my_validate_func()">
            <input type="text" name="phone" id="phone" onchange="my_validate_func()">
            <input type="text" name="course" id="course" onchange="my_validate_func()">
            <input type="text" name="center" id="center" onchange="my_validate_func()">
            <input type="text" name="roll" id="roll" value="">
        </form>
    </body>
    
    </html>
    
    **submit.php code is below**
    
    <?php
    $roll=rand(100000,999999);
    echo $roll;
    ?>
    
    
    ** Updated submit.php code is below**
    
    <?php
    

    Function getUniqueRandomNo() { $roll=rand(100000,999999); // Check in db, for example in emp_id field $id = mysql_query("select emp_id from emp_master where emp_id = '$roll'"); If($id !=null) { return getUniqueRandomNo(); } else{

    Return $roll; } } echo getUniqueRandomNo(); ?> Note: I have written this recursive function code directly here, please do necessary update. There might be syntax error.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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