Question

请查看数据：{limit: lim , offset: off, personname: $personname, fname: $fname},

这是将数据从当前php传递到ajax的正确方法吗？

<?php
$personname='Prakash';
$fname='';
$surname='';
?>
<script> 
function displayRecords(lim, off) {
    $.ajax({
        type: "GET",
        async: false,
        url: "getrecords.php",
        data: {personname: $personname, fname: $fname},
        success: function(html) {
            $("#results").append(html);
                                }
          });
}
</script>

Answer 1

在你的代码中，你必须这样写：

php variable中的

JavaScript应写为'<?php echo $val;?>'

 <?php 
    $personname='Prakash';
    $fname='';
    $surname='';
   ?>
    function displayRecords(lim, off) {
    $.ajax({
            type: "GET",
            async: false,
            url: "getrecords.php",
            data: {limit: lim , offset: off, personname: '<?php echo $personname;?>, fname: '<?php echo $fname?>'},
            beforeSend: function() {
            $("#loader_message").html("").hide();
            $('#loader_image').show();
            },
            success: function(html) {
            $("#results").append(html);
            $('#loader_image').hide();
            }
            });
            }

Answer 2

<?php
    $personname='Prakash';
    $fname='';
    $surname='';
?>
<script>
    function displayRecords(lim, off) {
            $.ajax({
                    type: "GET",
                    async: false,
                    url: "getrecords.php",
                    data: {limit: lim , offset: off, personname: <?php echo $personname;?>, fname:<?php echo $fname;?>},
                    beforeSend: function() {
                        $("#loader_message").html("").hide();
                        $('#loader_image').show();
                    },
                    success: function(html) {
                        $("#results").append(html);
                        $('#loader_image').hide();
                    }
                    });
            }

</script>

Answer 3

PHP variable passed应该jquery，如下所示

data: {personname: '<?php echo $personname; ?>', fname: '<?php echo $fname; ?>'},

将当前php页面中的变量数据传递给ajax然后传递给其他php

3 个答案: