Question

我创建一个表单（form1）来搜索mysql数据到表中而无需使用ajax重新加载页面，它工作正常，但是当我想从第二个表单（formorder）插入数据时，提交按钮不刷新或重新加载。

这是我的代码

的index.php

<html>
<head>
    <title>Search Data Without Page Refresh</title>
    <script type='text/javascript' src='js/jquery-1.4.2.min.js'></script>

</head>
<body>
<center>
<br>
<?php include("koneksi1.php");
$sql="select * from supplier";
$query=mysqli_query($koneksi,$sql);

 ?>
<form action="" name = "form1"> 
    <select name="namea" style="width:300px; padding:8px;">
        <?php 
            while($data=mysqli_fetch_assoc($query)){
                echo "<option value='".$data['id_supplier']."'>".$data['id_supplier']."</option>";
            }
         ?>
    </select>
    <input type="submit" value="Search" id="search-btn" style="padding:8px;"/>
</form>
<br>

<div id = "s-results">
    <!-- Search results here! -->
</div>

<script type = "text/javascript">
$(document).ready(function(){
    $('#s-results').load('search.php').show();


    $('#search-btn').click(function(){
        showValues();
    });

    $(function() {
        $('form').bind('submit',function(){
            showValues(); 
            return false; 
        });
    });

    function showValues() {

        $.post('search.php', { namea: form1.namea.value },

        function(result){
            $('#s-results').html(result).show();
        });
    }

});
</script>

</center>
</body>
</html>

的search.php

<?php
include("koneksi1.php");
isset( $_REQUEST['namea'] ) ? $name=$_REQUEST['namea'] : $name='';

$name = mysqli_real_escape_string($koneksi,$name);

if( empty( $name )){
echo '<script> alert("Please search something!")</script>';
}else{
    $sql = "select * from barang where id_supplier like '%$name%'";

    $rs = mysqli_query($koneksi,$sql ) or die('Database Error: ' . mysql_error());
    $num = mysqli_num_rows($rs);

    if($num >= 1 ){
    echo "<div style='margin:10px; color:green; font-weight: bold;'>$num records found!</div>";
    echo "<table width='365' border ='0' cellspacing='5' cellpadding='5'>";
    echo"<tr><th>RESULTS</th></tr>";
    echo "<div class='table-responsive'>";
    echo "<table class='table table-hover'>";

    while($data = mysqli_fetch_array( $rs )){

            ?>

        <tbody>
            <?php
            $kuantitas=0;
            $kuantitas++;
            ?>
            <form action="aksi.php" method="post" id="formorder">
            <tr>
                <td><?php echo $data['nama_barang']?></td>
                <td><?php echo $data['stok']?></td>
                <td><input type="text" name="kuantitas" size="4"></td>
                <td><input type="text" name="harga_satuan" id="harga_satuan" size="10"></td>
                <td><input type="text" name="ppn" id="ppn" size="3"> %</td>
                <td><button type="submit" class="btn btn-success btn-sm" id="add" name="add">ADD</button></td>
            </tr>
            </form>
            <?php }?>
        </tbody>
    </table>
</div>
    <?php  
    }else{
        echo "<tr><th>NO RESULTS</th></tr>";
    }
}
?>

按钮提交 formorder 无效，我的代码有什么问题???

提交表格不工作/不重新加载

0 个答案: