PHP插入建议到SQL表中
我不知道什么是错误的,目前没有错误,当我去我的建议页,所以我不知道什么是错的,有什么想法?PHP插入建议到SQL表中
//index.php
<form name="suggestion" action="http://example.com/suggestions.php" method="POST">
<font style="font-family: arial; color:#3f3f3f; font-size:15px;">Provide me with a URL to one or multiple songs that you want on the website!</font> <br><br><input style="width:243px;" type="text" name="suggestion"><br>
<br>
<input value="Submit" type="submit">
</form>
//suggestions.php
<?php
$conn=mysql_connect("localhost", "u611142741_list", "[REDACTED]");
mysql_select_db("u611142741_sugge", $conn);
// If the form has been submitted
$suggestion = mysql_real_escape_string($_POST['suggestion']);
$ip = $_SERVER['REMOTE_ADDR'];
// Build an sql statment to add the student details
$sql="INSERT INTO suggestions
(Suggestion,IP Address) VALUES
('$suggestion','$ip')";
$result = mysql_query($sql,$conn);
// close connection
mysql_close($conn);
?>
您的字段名称中有一个空格,并且需要加引号。你现在拥有什么;
INSERT INTO suggestions (Suggestion,IP Address) VALUES ('$suggestion','$ip')
...但字段名IP Address
包含空格,需要使用反引号,导致在MySQL中被引用,正常;
INSERT INTO suggestions (Suggestion, `IP Address`) VALUES ('$suggestion','$ip')
好吧,现在属性出现在我的sql表中,但建议是空白的,你知道什么是错的吗? – user4191887 2014-11-20 19:05:34
@ user4191887难道你的整个表单的命名与你的文本字段是一样的吗? – 2014-11-20 19:07:26
Errm ..我已将表单的名称更改为“建议”,但它仍未出现。 – user4191887 2014-11-20 19:37:11
你可以试试这个:
$sql="INSERT INTO suggestions
(`Suggestion`,`IP Address`) VALUES
('$suggestion','$ip')";
让我知道,如果作品。
您的字段名称是否真的带有空格的“IP地址”? – 2014-11-20 18:50:11
确实。 “IP地址” – user4191887 2014-11-20 18:51:28