準備在PHP語句不行

Question

嗨，我得到與後methos的值，我想與功能func_check_seven_userid（）

當我用它來檢查它：

$stmt = $this->conn->prepare("SELECT * FROM content where content_id= ".$this->seven); 
$stmt->execute(); 

它的工作。

但是當我使用：

$stmt = $this->conn->prepare("SELECT * FROM content WHERE content_id = :id"); 
    $stmt->execute(array(":id" => $this->seven)); 

實在不行！

我的完滿成功的代碼是：

感謝您的幫助

Answer 1

我發現這一點：

<?php 
/* Execute a prepared statement by passing an array of values */ 
$sql = 'SELECT name, colour, calories 
FROM fruit 
WHERE calories < :calories AND colour = :colour'; 
$sth = $dbh->prepare($sql, array(PDO::ATTR_CURSOR => PDO::CURSOR_FWDONLY)); 
$sth->execute(array(':calories' => 150, ':colour' => 'red')); 
$red = $sth->fetchAll(); 
$sth->execute(array(':calories' => 175, ':colour' => 'yellow')); 
$yellow = $sth->fetchAll(); 
?> 

上http://php.net/manual/en/pdo.prepare.php

這裏的語句與PDO::ATTR_CURSOR => PDO::CURSOR_FWDONLY

準備

也許它有區別。希望它有幫助