Question

我正在尝试从ajax调用子页面

<script type="text/javascript">
      function search(){
        //now lets fetch value of every input here
        var roll=$('#roll').val();
        // ctrl+shift+d
        var stu_name=$('#stu_name').val();
        var sta_name=$('#sta_name').val();
        var stu_class=$('#stu_class').val();
        var stu_batch=$('#stu_batch').val();
        var i_from=$('#i_from').val();
        var i_to=$('#i_to').val();
        // alert(i_to);
        $.ajax({
            url:root_path+"module/stationary/search.php",
            data:"roll="+roll+"&stu_name="+stu_name+"&sta_name="+sta_name+"&stu_class="+stu_class+"&stu_batch="+stu_batch+"&i_from="+i_from+"&i_to="+i_to,
            type:"post",
            success:function(e){
                $('#show').html(e);
                // alert(e);
            },error:function(e){
                alert(e);
            }
        });
      }
  </script>

它向我显示了无法在控制台中的ajax错误上读取null的属性'childNodes'

<?php
        $phpfiles=glob("../../include/php/*.php");
        foreach ($phpfiles as  $phpfile) {
            include_once($phpfile);
        }
        //boreder not comming properly
    ?>
    <tr class="active1 table_style" style="font-weight:600;">
        <td width="50">Sr. No.</td>
        <td width="50">Roll No.</td>
        <td width="150">Student Name</td>
        <td width="100">Father Name</td>
        <td width="100">Amount</td>
        <td width="100">Date</td>
        <td width="100">Stationary Name</td>
        <!-- <td width="50">Bounced Cheque</td>
        <td width="50">Cleared Cheque</td>
        <td width="50">Expired Cheque</td>
        <td width="50">Status</td>
        <td width="200" colspan="3">Action</td> -->
    </tr>
    <?php
    extract($_POST);
    //to select all data
    $wh=" where 1 ";
    if($stu_name!=''){
        $wh.=" and stu_profile.name like '%$stu_name%'";
    } 
    if($sta_name!=''){
        $wh.=" and stationary.name like '%$sta_name%'";
    } 
    if( $stu_class!=''){
        $wh.=" and class_id=$stu_class";
    } 
    elseif( $stu_class==''){
        $wh.=" and class_id!=''";
    } 
    if( $stu_batch!=''){
        $wh.=" and batch_id=$stu_batch";
    } 
    if($roll!=''){
        $wh.=" and (stationary.roll_no like '$roll%')";
    } 
    if($i_from!='' && $i_to==''){
        $wh.=" and datee='$i_from'";
    }
    elseif($i_from=='' && $i_to!=''){
        $wh.=" and datee='$i_to'";
    }elseif($i_from!='' && $i_to!=''){
        $wh.=" and (datee>='$i_from' and datee<='$i_to')";
    }
        $qry="select stationary.id,stu_profile.name as student , stationary.name,amount,datee,stationary.roll_no,stu_profile.name as stu,f_name from stationary join stu_profile on stationary.roll_no=stu_profile.roll_no $wh";
        echo "<tr><td>$qry</td></tr>";
        $pdc_dtl=fetchAll($qry);
        // print_r($pdc_dtl);
        $sno=1;
        foreach ($pdc_dtl as  $pdc_value) {
            ?>
    <tr class="primary1 table_style2">
        <td><?php echo $sno++; ?></td>
        <td><?php echo $pdc_value['roll_no'] ?></td>
        <td><?php echo $pdc_value['student'] ?></td>
        <td><?php echo $pdc_value['f_name'] ?></td>
        <td><?php echo $pdc_value['amount'] ?></td>
        <td><?php echo $pdc_value['datee'] ?></td>
        <td><?php echo $pdc_value['name'] ?></td>

    </tr>
    </table>
    <?php } ?>

这是search.php文件当我警告javascript变量其显示正确的答案时，我需要根据搜索显示数据，并在我写了4位数字以滚动其显示查询后调用搜索功能onkeyup

无法在Ajax上读取null的属性“ childNodes”

0 个答案: