我很难弄清楚如何使用 php 将选中的框或未选中的框插入到数据库中。
我尝试了很多不同的方法，但都没有用，我想我已经很接近了，但无法解决问题。 顺便说一句，我使用 Javascript，除了这次以外从未使用过 PHP。
index.html
<html>
<head>
<style type="text/css">
@import "demo_page.css";
@import "header.ccss";
@import "demo_table.css";
@import "select.dataTables.min.css";
@import "jquery.dataTables.min.css";
</style>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script type="text/javascript" charset="utf-8" src="jquery.js"></script>
<script type="text/javascript" charset="utf-8" src="jquery.dataTables.js"></script>
<script type="text/javascript" charset="utf-8" src="RowGroupingWithFixedColumn.js"></script>
<script>$(document).ready(function(){load_(); console.log('load running')});</script>
</head>
<body id="dt_example">
<table cellpadding="0" cellspacing="0" border="0" class="display" id="endpoints">
<thead>
<tr>
<th></th>
<th>Nr.</th>
<th>Java Class Name</th>
<th>http Method</th>
<th>URL</th>
</tr>
</thead>
<tbody>
<?php
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,DB_NAME);
$sql='SELECT * FROM url';
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
print $row['method'];
switch ($row['http_method']) {
case "GET":
echo "<tr class='gradeA'>";
break;
case "PUT":
echo "<tr class='gradeC'>";
break;
case "POST":
echo "<tr class='gradeU'>";
break;
case "DELETE":
echo "<tr class='gradeX'>";
break;
default:
echo "<tr>";
}
if($row['checked']){
echo "<td><input type='checkbox' id=case name='case[]' value='" . $row['number'] . "' checked> </td>";
} else {
echo "<td><input type='checkbox' id=case name='case[]' value='" . $row['number'] . "'> </td>";
}
echo "<td align=center >" . $row['number'] . "</td>";
echo "<td align=center >" . $row['class_name'] . "</td>";
echo "<td>" . $row['http_method'] . "</td>";
echo "<td style='font-weight:bold'>" . $row['endpoint'] . "</td>";
echo "</tr>";
}
if(isset($_POST['save'])){
echo "<script>console.log(" . $checkboxes.length . ");</script>";
$rows = $_POST['case'];
foreach($rows as $row){
$sql = "UPDATE url SET checked = 1 WHERE number = " . $case;
$result = mysqli_query($con,$sql);
}
}
mysqli_close($con);
echo "</tbody></table>";
echo "<input type='submit' name='save' id='save' value='Save' />";
?>
</body>
</html>
如有任何帮助，我们将不胜感激。
图片:
编辑: 已解决 - 这是代码:
<?php
if(isset($_POST['save'])){
echo "<script>console.log(" . $checkboxes.length . ");</script>";
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,DB_NAME);
$rows = $_POST['case'];
foreach($rows as $row){
$sql = "UPDATE url SET checked = 1 WHERE number = " . $row;
$result = mysqli_query($con,$sql);
}
}
?>
<!DOCTYPE html>
<html>
<head>
<style type="text/css">
@import "demo_page.css";
@import "header.ccss";
@import "demo_table.css";
@import "select.dataTables.min.css";
@import "jquery.dataTables.min.css";
</style>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script type="text/javascript" charset="utf-8" src="jquery.js"></script>
<script type="text/javascript" charset="utf-8" src="jquery.dataTables.js"></script>
<script type="text/javascript" charset="utf-8" src="RowGroupingWithFixedColumn.js"></script>
<script>$(document).ready(function(){load_(); console.log('load running')});</script>
</head>
<body id="dt_example">
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<table cellpadding="0" cellspacing="0" border="0" class="display" id="endpoints">
<thead>
<tr>
<th></th>
<th>Nr.</th>
<th>Java Class Name</th>
<th>http Method</th>
<th>URL</th>
</tr>
</thead>
<tbody>
<?php
$con = mysqli_connect('sql7.freemysqlhosting.net','sql7117068','GZqaZj69G9','sql7117068');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,'sql7117068');
$sql='SELECT * FROM url';
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
print $row['method'];
switch ($row['http_method']) {
case "GET":
echo "<tr class='gradeA'>";
break;
case "PUT":
echo "<tr class='gradeC'>";
break;
case "POST":
echo "<tr class='gradeU'>";
break;
case "DELETE":
echo "<tr class='gradeX'>";
break;
default:
echo "<tr>";
}
if($row['checked']){
echo "<td><input type='checkbox' id=case name='case[]' value='" . $row['number'] . "' checked> </td>";
} else {
echo "<td><input type='checkbox' id=case name='case[]' value='" . $row['number'] . "'> </td>";
}
echo "<td align=center >" . $row['number'] . "</td>";
echo "<td align=center >" . $row['class_name'] . "</td>";
echo "<td>" . $row['http_method'] . "</td>";
echo "<td style='font-weight:bold'>" . $row['endpoint'] . "</td>";
echo "</tr>";
}
mysqli_close($con);
echo "</tbody></table>";
echo "<input type='submit' name='save' id='save' value='Save' />";
echo "</form>";
?>
</body>
</html>
PS:代码中可能有很多漏洞和糟糕的编程风格，但要知道我不是 PHP 开发人员，如果我的服务器受到不同的攻击也没关系。我只想解决问题并继续前进。没有时间在每一点都停下来。
最佳答案
使用下面的代码检查复选框是否被选中。然后在一些变量上设置一些标志以相应地将数据插入到数据库中。
//html
<input type='checkbox' name='boxname' value='1' />
//php
<?php
if(isset($_POST['boxname'])
{
echo "check box is checked";
}
?>
关于php - 使用html php POST选中和未选中的复选框，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36936123/