html表单tp php文件

时间:2017-08-27 07:34:13

标签: php html mysql

我正在尝试构建一个html表单并将其连接到正在执行查询并将信息插入mysql数据库的php文件。

这是我的HTML代码:

<!DOCTYPE html>

<html lang="en">

<head>

<meta charset="UTF-8">

<title>Add Record Form</title>

</head>

<body>

<form action="/joomla_31/insert.php" method="post">


<label for="MGMT_IP">MGMT IP</label>

<input type="text" name="MGMT_IP" id="MGMT_IP">

</p>

<p>

<label for="Vendor">Vendor</label>

<input type="text" name="Vendor" id="Vendor">

</p>

<p>

<label for="Version">Version</label>

<input type="text" name="Version" id="Version">

</p>

<p>

<label for="GUI_User">GUI User</label>

<input type="text" name="GUI_User" id="GUI_User">

</p>

<p>

<label for="GUI_Pass">GUI Pass</label>

<input type="text" name="GUI_Pass" id="GUI_Pass">

</p>


<input type="submit" value="submit">

</form>

</body>

</html>

这是我的insert.php文件:

<?php


$link = mysqli_connect("localhost", "root", "", "mysql");



// Check connection

if($link === false){

die("ERROR: Could not connect. " . mysqli_connect_error());

}



// Escape user inputs for security

$MGMT_IP = mysqli_real_escape_string($link, $_REQUEST['MGMT IP']);
$Vendor = mysqli_real_escape_string($link, $_REQUEST['Vendor']);
$Version = mysqli_real_escape_string($link, $_REQUEST['Version']);
$GUI_User = mysqli_real_escape_string($link, $_REQUEST['GUI User']);
$GUI_Pass = mysqli_real_escape_string($link, $_REQUEST['GUI Pass']);



// attempt insert query execution

$sql = "INSERT INTO `cloud_team` (MGMT IP, Vendor, Version, GUI User, GUI Pass) VALUES (`$customer_number`, `$customer_name`, `$MGMT_IP`, `$Vendor`, `$Version`, `$GUI_User`, `$GUI_Pass`)";

if(mysqli_query($link, $sql)){

echo "Records added successfully.";

} else{

echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);

}


// close connection

mysqli_close($link);

?>

但是每次点击提交按钮我都会得到一个空白页面...... 这就像重定向正确的文件(http://x.x.x.x/joomla_31/insert.php)..但没有执行...没有显示错误..只有空白页

1 个答案:

答案 0 :(得分:0)

启用PHP错误显示以更好地调试代码。有关详细信息,请查看this答案。