2015-01-26 82 views
0

试图保存表两行的时候不工作:victimstmt->执行插入两行

假设的输出是这样的:

name | age | gender | address   | Journal_id 
---------------------------------------------------------------- 
    robert | 23 | M  | st. lucas calabanga |  1 
    vender | 25 | M  | st. lucas calabanga |  1 

我写了一个准备语句多种绑定PARAM这里是:

$j_id = mysqli_insert_id($this->db); 

    $stmt = $this->db->prepare("INSERT INTO victim (name, age, gender, address, journal_id) VALUES (?, ?, ?, ?, ?)"); 

    $length = count($vName); 
    for ($i = 0; $i < $length; $i++) 
    { 
     if ($stmt->bind_param('sissi', $vName[$i], $vAge[$i], $vGender[$i], $vAddr[$i], $j_id)) 

      if($stmt->execute()) 
       echo "working"; 
      else 
       echo "not working"; 
    } 

当我运行该代码时,它没有在我的数据库惰性。任何能够帮助我的人。

+0

我想你错过了这个链接:http://stackoverflow.com/help/formatting – Rizier123 2015-01-26 19:55:21

+0

“没有插入”**如何**?你有“不工作”? – 2015-01-26 20:01:27

+0

@MarcB是的,不工作... – 2015-01-26 20:07:16

回答

0
for ($i = 0; $i < $length; $i++) { 
    if($stmt->execute(array($vName[$i], $vAge[$i], $vGender[$i], $vAddr[$i], $j_id))) { 
     echo "working"; 
    } 
    else { 
     echo "not working"; 
    } 
}