我的php代码没有更新或添加到数据库
基本上，我没有得到任何错误，当我按下更新或添加。有人能帮我吗？我需要它，所以当我按add时，它会将数据添加到数据库中，当我按update时，它会用该值更新数据库我的php代码没有更新或添加到数据库,php,Php,基本上，我没有得到任何错误，当我按下更新或添加。有人能帮我吗？我需要它，所以当我按add时，它会将数据添加到数据库中，当我按update时，它会用该值更新数据库 <html> <head> <title>Subcontractors Data</title> </head> <body> <a href="index.html">Logout</a> <a href="homepage.htm
<html>
<head>
<title>Subcontractors Data</title>
</head>
<body>
<a href="index.html">Logout</a>
<a href="homepage.html">Homepage</a>
<?php
//make connection
$con = mysqli_connect("localhost","root","");
if(!$con){
die("Can not connect " . mysqli_error());
}
//select db
mysqli_select_db($con , 'subcontractor');
$sql="SELECT * FROM subcontractors";
if(isset($_POST['update'])){
$UpdateQuery = "UPDATE subcontractors SET ID='$_POST[ID]', Name='$_POST[Name]', Surname='$_POST[Surname]', FPA='$_POST[FPA]', Performance='$_POST[Performance]' WHERE ID='$_POST[hidden]'";
mysqli_query($con, $UpdateQuery);
};
if(isset($_POST['add'])){
$AddQuery = "INSERT INTO subcontractors (ID, Name, Surname, FPA, Performance) VALUES ('$_POST[aID]','$_POST[aName]','$_POST[aSurname]','$_POST[aFPA]','$_POST[aPerformance]')";
mysqli_query($con, $AddQuery);
};
$my_Data=mysqli_query($con,$sql);
echo "<table border=1>";
echo"<tr>";
echo"<th>ID</th>";
echo"<th>Name</th>";
echo"<th>Surname</th>";
echo"<th>FPA</th>";
echo "<th>Performance</th>";
echo "</tr>";
while($record=mysqli_fetch_assoc($my_Data)){
echo "<form action=editsub.php method=post>";
echo "<tr>";
echo "<td>" . "<input type=text name='ID' value=".$record['ID'] ." </td>";
echo "<td>" . "<input type=text name='Name' value=".$record['Name'] . " </td>";
echo "<td>" . "<input type=text name='Surname' value=".$record['Surname'] . " </td>";
echo "<td>" . "<input type=text name='FPA' value=".$record['FPA'] . "% </td>";
echo "<td>" . "<input type=text name='Performance' value=".$record['Performance'] . "% </td>";
echo "<input type=hidden name='hidden' value=" . $record['ID'] . ">";
echo "<input type=submit name='update' value='update'>";
echo "</tr>";
echo "</form>";
}
echo "<form action=editsub.php mehtod=post>";
echo "<tr>";
echo "<td><input type=text name='aID'></td>";
echo "<td><input type=text name='aName'></td>";
echo "<td><input type=text name='aSurname'></td>";
echo "<td><input type=text name='aFPA'></td>";
echo "<td><input type=text name='aPerformance'></td>";
echo "<td>" . "<input type=submit name='add' value='add'" . " </td>";
echo "</form>";
echo "</table>";
mysqli_close($con);
?>
</body>
</html>
分包商数据
这样您就可以检查错误
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
第一种，方法
！=<代码>方法editsub.php