PHP 将表单提交到另一个页面，运行方法

Question

我有 2 个 php 文件：a.php 和 b.php

我想提交从 a.php 到 b.php 的数据。但我试过了，不是 b.php 中的运行。（我想在 a.php 中输入 b.php 中的运行 "search" 方法）。

谢谢

这是我的代码

a.php

<form action="b.php" name="form1" method="post" >
<input type="search" name="search" id="search" >
<input type="submit" value=" "></form>

b.php

<form id="form1" name="form1" method="post" action="" >
<input type="search" value="<?php echo $_POST['search'];?>" name="search">
<input type="submit" name="cari" value=" "></form>
<?php $search = $_POST['search']; $cari = $_POST['cari']; if($cari){
if($search != "") { $query = "SELECT  * FROM kondef where poin
like'%$search%' or deskripsi like '%$search%'"; $result =   $db>
query($query);}  else { die (mysql_error()); }while($row = $result- 
fetchArray()) {?><tr>   <td><?php echo $row['poin'];?></td><td><?php echo 
$row['deskripsi'];?></td><td><?php echo $row['link'];?></td></tr><?php } ?>
</table>

Answer 1

尝试使用下面的代码

a.php

<form action="b.php" name="form1" method="post" >
   <input type="search" name="search" id="search" >
   <input type="submit" name="cari" value="submit">
</form>

b.php

<form id="form1" name="form1" method="post" action="" >
    <input type="search" value="<?php echo $_POST['search'];?>" name="search">
    <input type="submit" name="cari" value="submit">
</form> 
<table>

<?php $search = $_POST['search'];
    if(($search != "") && ($_POST['cari'] != "")) { 
        $query = "SELECT * FROM kondef WHERE poin like '%$search%' or deskripsi like '%$search%'";
        $result = $db->query($query);
    } else {
        die (mysql_error()); 
    };
    while($row = $result->fetchArray()) {?>
    <tr>
        <td><?php echo $row['poin'];?></td>
        <td><?php echo $row['deskripsi'];?></td>
        <td><?php echo $row['link'];?></td>
    </tr>
    <?php } ?>
</table>

PHP 将表单提交到另一个页面，运行 方法

PHP submit form to another page, and run the method

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submit

PHP 将表单提交到另一个页面，运行方法