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기본적으로 업데이트 또는 추가를 누르면 오류가 표시되지 않습니다. 누구든지 나를 도울 수 있습니까? 나는 그것을 추가 할 때 데이터베이스에 데이터를 추가하고, update를 누르면 해당 값으로 데이터베이스를 업데이트하기를 원합니다.PHP 코드가 업데이트 중이거나 데이터베이스에 추가되지 않았습니다.
<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>
처음으로,'mehtod'! ='method'. 둘째, 준비된 문장을 사용하십시오. 주입 공격에 개방적입니다. 셋째,이 'editsub.php'입니까? 넷째, 질의가 실제로 실행되고 있는지 확인하고, 마지막으로 [오류]를 확인하십시오 (http://php.net/manual/en/mysqli.error.php). –