mysqli準備聲明錯誤？

Question

$ mysqli = new mysqli（「localhost」，「root」，「」，「test」）; $ mysqli-> query（'PREPARE mid FROM「SELECT name FROM test_user WHERE id =？」'）; //工作代碼開始 // $ res = $ mysqli-> query（'PREPARE mid FROM FROM SELECT name FROM test_user''）; // $ res = $ mysqli-> query（'EXECUTE mid;'）or die（mysqli_error（$ mysqli））; //工作代碼結束.. $ res = $ mysqli-> query（'EXECUTE mid 1;'）or die（mysqli_error（$ mysqli））;

while($resu = $res->fetch_object()) { 
    echo '<br>' .$resu->name; 
} 

錯誤：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 

我的php版本是PHP Version 5.3.0和MySQL

mysqlnd 5.0.5-dev - 081106 - $Revision: 1.3.2.27 $ 

我使用where clause

Answer 1

使用了該準備的功能得到了與出正確的結果SELECT查詢：

http://php.net/manual/en/mysqli.prepare.php

/* create a prepared statement */ 
if ($stmt = $mysqli->prepare("SELECT District FROM City WHERE Name=?")) { 

    /* bind parameters for markers */ 
    $stmt->bind_param("s", $city); 

    /* execute query */ 
    $stmt->execute(); 

    /* bind result variables */ 
    $stmt->bind_result($district); 

    /* fetch value */ 
    $stmt->fetch(); 

    /* close statement */ 
    $stmt->close(); 
}

Answer 2

http://dev.mysql.com/doc/refman/5.1/en/execute.html

您必須在EXECUTE語句使用條款。所以你可以做這樣的事情：

$mysqli = new mysqli("localhost", "root", "", "test"); 
$mysqli->query('PREPARE mid FROM "SELECT name FROM test_user WHERE id = ?"'); 
$mysqli->query('SET @var1 = 1;'); 
$res = $mysqli->query('EXECUTE mid USING @var1;') or die(mysqli_error($mysqli));