PHP插入建議到SQL表中

Question

我不知道什麼是錯誤的，目前沒有錯誤，當我去我的建議頁，所以我不知道什麼是錯的，有什麼想法？

//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

您的字段名稱中有一個空格，並且需要加引號。你現在擁有什麼;

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

...但字段名IP Address包含空格，需要使用反引號，導致在MySQL中被引用，正常;

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

Answer 2

你可以試試這個：

$sql="INSERT INTO suggestions 

(`Suggestion`,`IP Address`) VALUES 

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

讓我知道，如果作品。