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php - 我的代码在搜索后不会更新

转载 作者:行者123 更新时间:2023-11-29 13:16:16 24 4
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任何人都可以帮我找到导致我的代码无法工作的代码吗?我的代码不会更新...我已经调试这个代码 3 个小时了,但仍然无法修复它:(...我需要你们的帮助。

PHP 代码:

<?php
if(isset($_GET['gogo'])){
include('include/connect.php');
$batchcode = $_GET['code'];
$sql = mysql_query("SELECT * FROM score WHERE batchcode = '".$batchcode."' ");
if($sql) {
while($rows = mysql_fetch_array($sql)){
$id[] = $rows['id'];
$name[] = $rows['name'];
$score1[] = $rows['score1'];
$score2[] = $rows['score2'];
$other_qual[] = $rows['score3'];
$interview[] = $rows['score4'];
$total[] = $rows['total'];
}
}
}
?>
<?php
if(isset($_POST['update'])){
include('include/connect.php');
//1
$u1id = $_POST['id1'];
$u1name = $_POST['name1'];
$u1score1 = $_POST['optA1'];
$u1score2 = $_POST['optB1'];
$u1other_qual = $_POST['other_qual1'];
$u1interview = $_POST['interview1'];
$u1total = $_POST['total1'];
//2
$u2id = $_POST['id2'];
$u2name = $_POST['name2'];
$u2score1 = $_POST['optA2'];
$u2score2 = $_POST['optB2'];
$u2other_qual = $_POST['other_qual2'];
$u2interview = $_POST['interview2'];
$u2total = $_POST['total2'];
//1
mysql_query("UPDATE score SET score1='$u1score1', score2='$u1score2', total='$u1total' WHERE id='$u1id'");
//2
mysql_query("UPDATE score SET score1='$u2score1', score2='$u2score2', total='$u2total' WHERE id='$u2id'");

header("Location: index.php");
}
?>

html代码:

<form method="get">
<form method="post">
Search batchcode: <input type="text" name="code" id="query" /><input type="submit" value="Go" name="gogo" /><br />
<table>
<tr>
<td>
ID: <br />
<input type="text" name="id1" value="<?php if(empty($id[0])){$id[0] = array(NULL);}else{echo $id[0];} ?>" readonly /> <br />
<input type="text" name="id2" value="<?php if(empty($id[1])){$id[1] = array(NULL);}else{echo $id[1];} ?>" readonly /> <br />
</td>
<td>
Name: <br />
<input type="text" name="name1" value="<?php if(empty($name[0])){$name[0] = array(NULL);}else{echo $name[0];} ?>" readonly /> <br />
<input type="text" name="name2" value="<?php if(empty($name[1])){$name[1] = array(NULL);}else{echo $name[1];} ?>" readonly /> <br />
</td>
<td>
Score 1: <br />
<input type="text" name="optA1" value="<?php if(empty($score1[0])){$score1[0] = array(NULL);}else{echo $score1[0];} ?>" onchange="optTotal1()" /> <br />
<input type="text" name="optA2" value="<?php if(empty($score1[1])){$score1[1] = array(NULL);}else{echo $score1[1];} ?>" onchange="optTotal2()" /> <br />
</td>
<td>
Score 2: <br />
<input type="text" name="optB1" value="<?php if(empty($score2[0])){$score2[0] = array(NULL);}else{echo $score2[0];} ?>" onchange="optTotal1()" /> <br />
<input type="text" name="optB2" value="<?php if(empty($score2[1])){$score2[1] = array(NULL);}else{echo $score2[1];} ?>" onchange="optTotal2()" /> <br />
</td>
<td>
Other Qualification: <br />
<input type="text" name="other_qual1" value="<?php if(empty($other_qual[0])){$other_qual[0] = array(NULL);}else{echo $other_qual[0];} ?>" readonly /> <br />
<input type="text" name="other_qual2" value="<?php if(empty($other_qual[1])){$other_qual[1] = array(NULL);}else{echo $other_qual[1];} ?>" readonly /> <br />
</td>
<td>
Interview: <br />
<input type="text" name="interview1" value="<?php if(empty($interview[0])){$interview[0] = array(NULL);}else{echo $interview[0];} ?>" readonly /> <br />
<input type="text" name="interview2" value="<?php if(empty($interview[1])){$interview[1] = array(NULL);}else{echo $interview[1];} ?>" readonly /> <br />
</td>
<td>
Total: <br />
<input type="text" name="total1" value="<?php if(empty($total[0])){$total[0] = array(NULL);}else{echo $total[0];} ?>" readonly onKeyUp="optTotal1()" /> <br />
<input type="text" name="total2" value="<?php if(empty($total[1])){$total[1] = array(NULL);}else{echo $total[1];} ?>" readonly onKeyUp="optTotal2()" /> <br />
</td>
</tr>
</table>
<input type="submit" value="update" name="update" />
</form>
</form>

最佳答案

您不能同时执行 GET 和 POST。使用其中之一..

在 HTML 中删除 <form method="get">以及相应的</form>只需使用 POST 即可。 (<form method="post">)

请参阅:Post and get at the same time in php

然后在你的 PHP 中,更改 GETPOST像这样:

if(isset($_POST['gogo'])){
include('include/connect.php');
$batchcode = $_POST['code'];
$sql = mysql_query("SELECT * FROM score WHERE batchcode = '".$batchcode."' ");
...

编辑:

或者,您可以保持您的 php 代码与您拥有的方式相同,并在 HTML 中将其设为 2 个单独的表单,.. 搜索表单使用 GET另一种形式使用 POST

所以 HTML 应该是这样的:

<form method="get">
Search batchcode: <input type="text" name="code" id="query" /><input type="submit" value="Go" name="gogo" /><br />
</form>

<form method="post">
<table>
<tr>
<td>
ID: <br />
<input type="text" name="id1" value="<?php if(empty($id[0])){$id[0] = array(NULL);}else{echo $id[0];} ?>" readonly /> <br />
<input type="text" name="id2" value="<?php if(empty($id[1])){$id[1] = array(NULL);}else{echo $id[1];} ?>" readonly /> <br />
</td>...
...
</form>

关于php - 我的代码在搜索后不会更新,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21419231/

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