Javascript 如何将此php代码选项(textarea字段)插入mysql数据库
嗨,伙计们,你们对这个密码有些了解。textarea已提交到数据库。因为我有一个connection.php,但它没有插入到数据库,请帮助我使用选项textarea。你能帮我查一下phpmyadmin SQL吗?我应该把tnx放在什么名称、类型等上Javascript 如何将此php代码选项(textarea字段)插入mysql数据库,javascript,php,html,mysql,forms,Javascript,Php,Html,Mysql,Forms,嗨,伙计们,你们对这个密码有些了解。textarea已提交到数据库。因为我有一个connection.php,但它没有插入到数据库,请帮助我使用选项textarea。你能帮我查一下phpmyadmin SQL吗?我应该把tnx放在什么名称、类型等上 select.html <html lang="en"> <title>NTF Catering Service</title> <meta charset="utf-8"&g
select.html
<html lang="en">
<title>NTF Catering Service</title>
<meta charset="utf-8">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"> </script>
<script src="js/js.1.js" type="text/javascript"></script>
</head>
<body>
<form action="create.php" method="post">
<select multiple="multiple" class="options" id="textarea">
<option value="item1">Item 1</option>
<option value="item2">Item 2</option>
<option value="item3">Item 3</option>
<option value="item4">Item 4</option>
<option value="item5">Item 5</option>
</select>
<button id="copy">Copy</button>
<button id="remove">Remove</button>
<select id="textarea2" multiple class="remove">
<input type="submit" name="submit" />
</form>
</select>
</html>
connection.php
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$db = "copy";
$conn = mysql_connect($dbhost,$dbuser,$dbpass);
mysql_select_db($db);
?>
submit.php
<?php
include 'connection.php';
$food1 = $_POST['food1'];
$food2 = $_POST['food2'];
$food3 = $_POST['food3'];
$food4 = $_POST['food4'];
$food5 = $_POST['food5'];
if(!$_POST['submit']) {
echo "please fill out the form";
header('Location: select.html');
} else {
$sql = "INSERT INTO remove(food1, food2, food3, food4, food5) VALUES ('".$food1."', '".$food2."', '".$food3."','".$food4."','".$food5."');";
mysql_query($link, $sql);
echo "User has been added!";
header('Location: select.html');
}
?>
select.html
NTF餐饮服务
项目1
项目2
项目3
项目4
项目5
复制
去除
connection.php
submit.php
请将您的插入到语句中,格式如下:
$sql = "INSERT INTO remove(food1, food2, food3, food4, food5) VALUES ($food1, $food2, $food3, $food4, $food5)";
您的操作错误,应该是submit.php
另外,请不要使用mysql.*
函数,因为它们已被弃用
改用mysqli
或PDO
。还要使用准备好的语句
,以防止SQL注入
connection.php的示例:
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$db = "copy";
$conn = mysqli_connect($dbhost,$dbuser,$dbpass,$db);
?>
submit.php的示例(带有准备好的语句):
看起来您将表单数据发布到了错误的文件中?应该发布到submit.php,而不是create.php
vssubmit.php
;-)另外,清理你的POST
值!Reguardingmysql\u connect
-读取您尚未转义输入字符串,因此如果任何输入包含引号或其他SQL保留字符,则查询将失败。容易破碎。另外,这使得它很容易被黑客攻击。另外,请不要使用mysql\u xxx()
函数,它们已经过时,不推荐使用。您应该改用较新的mysqli
或PDO
库。@daniel mensing yah已将其更改为sry;)但仍然没有插入到database@Spudley好的,tnx:)在phpmyadmin SQL中,未正确使用的名称=食物类型=文本??取决于您的功能。但是,是的,文本是一种可能性。是的,那么你可以使用文本。如果你有很长的文本,你也可以使用长文本,但我相信文本是可以的。你能看看你的食物变量吗?回显所有$food变量并查看输出isi的内容,该代码用于html,foodA foodB foodC foodD foodD Copy Remove Change mysql\u查询($link,$sql)
到mysql_查询($link,$sql)或死亡(mysql_error())
查看您可能遇到的sql错误
<?php
include 'connection.php';
$food1 = $_POST['food1'];
$food2 = $_POST['food2'];
$food3 = $_POST['food3'];
$food4 = $_POST['food4'];
$food5 = $_POST['food5'];
if(!$_POST['submit']) {
echo "please fill out the form";
header('Location: select.html');
} else {
$sql = "INSERT INTO remove(food1, food2, food3, food4, food5) VALUES (?,?,?,?,?);";
$stmt = mysqli_prepare($conn, $sql);
mysqli_stmt_bind_param($stmt,"sssss",$food1,$food2,$food3,$food4,$food5);
mysqli_stmt_execute($stmt);
echo "User has been added!";
header('Location: select.html');
}
?>