Mysql的像bind_param（）非對象</p> <p>在非對象

Question

$id='gsf'; 
$stmt=$mysqli->prepare('SELECT `title` FROM `post` WHERE (`content` LIKE `%?%`) and accepted=0 order by count desc Limit 0,5'); 
     $stmt->bind_param('s',$id); 
     $stmt->execute(); 
     $stmt->bind_result($title); 
     while($stmt->fetch();) 
    { 
     echo $title."<br>";   
     } 
     $stmt->close(); 
     $mysqli->close(); 

這是一個PHP代碼段母雞我執行它，我得到這個錯誤調用一個成員函數bind_param（）對我失去了什麼出？

Answer 1

試試這個：

$stmt=$mysqli->prepare('SELECT `title` FROM `post` WHERE (`content` LIKE ?) and accepted=0 order by count desc Limit 0,5'); 

$stmt->bind_param('s',$id); 

$id = '%gsf%'; //The **%** go here 

$stmt->execute(); 
$stmt->bind_result($title); 

while($stmt->fetch()) { 
     echo $title."<br>";   
} 

$stmt->close(); 
$mysqli->close(); 

Answer 2

你的SQL是erronous，從而$stmt不是一份準備好的聲明。分析$mysqli->errno和$mysqli->error的內容以在您的SQL查詢中找到問題。