这是我的表格
<html>
<head><title>Hawkins Car Records</title></head>
<body><h1>Add New Car</h1></body>
<form action="carNewBack.php" method="POST">
Car Name: <input type="text" name="carName"/>
<br>
Make: <input type="text" name="make"/>
<br>
Model: <input type="text" name="model"/>
<br>
Year: <input type="text" name="year"/>
<br>
Last 5 digits of VIN: <input type="text" name="lastVIN"/>
<br>
Plate: <input type="text" name="plate"/>
<br><br>
<input type="submit" value="Submit"/>
</form>
</html>
当我点击提交按钮时,没有任何反应.没有白屏,没有404,没什么.它不会执行carNewBack.php.有人可以分享任何想法吗？
这是动作文件.我试图建立一个家庭汽车服务记录的数据库,这是一个输入并创建新车记录的形式.
<?php
$carconnect = mysqli_connect("localhost", "carUser", "caps271:snows", "cars");
if (mysqli_connect_errno()) {
printf("connect failed: %s\n", mysqli_connect_error());
exit();
} else {
$carName = mysqli_real_escape_string($_POST['carName']);
$make = mysqli_real_escape_string($_POST['make']);
$model = mysqli_real_escape_string($_POST['model']);
$year = mysqli_real_escape_string($_POST['year']);
$lastVIN = mysqli_real_escape_string($_POST['lastVIN']);
$plate = mysqli_real_escaped_string($_POST['plate']);
$sql = "INSERT INTO cars (carName, make, model, year, lastVIN, plate) VALUES ('". $carName."', '".$make."', '".$model."', '".$year."', '".$lastVIN."', '". $plate."')";
$res = mysqli_query($carconnect, $sql);
if ($res === TRUE) {
echo "Car added";
} else {
printf ("Could not insert car: %s\n", mysqli_error($carconnect));
}
mysqli_close($carconnect);
}
?>
编辑:代码修复.
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