Question

我不知道出了什么问题，进入我的建议页面时目前没有错误，所以我不知道出了什么问题，有什么想法吗？

//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);
?>

Answer 1

您的字段名称中有一个空格，需要引用它。你现在拥有的是什么;

INSERT INTO suggestions (Suggestion,IP Address) VALUES ('$suggestion','$ip')

...但字段名IP Address包含一个空格，需要引用，通常在MySQL中使用反引号，导致;

INSERT INTO suggestions (Suggestion, `IP Address`) VALUES ('$suggestion','$ip')

Answer 2

你可以试试这个：

$sql="INSERT INTO suggestions

(`Suggestion`,`IP Address`) VALUES

('$suggestion','$ip')";

如果有效，请告诉我。

PHP将Suggestion插入SQL表

2 个答案: