執行基於變量的sqlite3更新通過php

Question

我試圖通過PHP腳本執行更新到Sqlite3數據庫，但我遇到了一對我不明白的錯誤。

PHP Warning: SQLite3::prepare(): Unable to prepare statement: 1, no such column: :column in C:\UwAmp\www\save.php on line 16, referer: http://localhost/Canvas.html

PHP Fatal error: Uncaught Error: Call to a member function bindParam() on boolean in C:\UwAmp\www\save.php:17\nStack trace:\n#0 {main}\n thrown in C:\UwAmp\www\save.php on line 17, referer: http://localhost/Canvas.html

這裏是我使用的PHP代碼的例子：

$url = $_POST['newURL']; 
$name = $_POST['saveName']; 
$index = $_POST['saveNum']; 
$user = $_POST['username']; 
$db = new SQLite3('Users.db'); 



$statement = $db->prepare("UPDATE Canvas_Saves SET ':column' = ':url' WHERE User = ':user'"); 
$statement->bindParam(':url', $url); 
$statement->bindParam(':user', $user); 
if($index ===1){ 
    $statement->bindParam(':column', 'Save1',SQLITE3_TEXT); 
} 
else if($index ===2){ 
    $statement->bindParam(':column', 'Save2',SQLITE3_TEXT); 
} 
else if($index ===3){ 
    $statement->bindParam(':column', 'Save3',SQLITE3_TEXT); 
} 
else if($index ===4){ 
    $statement->bindParam(':column', 'Save4',SQLITE3_TEXT); 
} 

$statement->execute(); 

第16行是$ DB->準備發言。 我試圖研究這個，但我沒有任何運氣。感謝您提供任何幫助。

Answer 1

您可以使用此

$url = $_POST['newURL']; 
$name = $_POST['saveName']; 
$index = $_POST['saveNum']; 
$user = $_POST['username']; 
$db = new SQLite3('Users.db'); 
$col=""; 
if($index ===1){ 
    $col="Save1"; 
} 
else if($index ===2){ 
    $col="Save2"; 
} 
else if($index ===3){ 
    $col="Save3"; 
} 
else if($index ===4){ 
    $col="Save4"; 
} 
$statement = $db->prepare("UPDATE Canvas_Saves SET ".$col."=:url WHERE User = :user "); 
$statement->bindParam(":url", $url); 
$statement->bindParam(":user", $user); 
$statement->execute();