Question

我尝试将记录插入mysql数据库

INSERT INTO tbl_messages
(msgTitle,msgMessage,msgDateScheduled,fkmsgAppId,msgImage,msgAction,fkmsgSerial) 
values
('essfd','fgf','2014-03-30T00:12','2','test.jpg','none','11111');

但是虽然声明是正确的，但我仍然收到这个错误：

查询无效：您的SQL语法出错;检查手册对应于您的MariaDB服务器版本以获得正确的语法使用附近＆＃39; ＆＃39;在第1行

如果我在mysql客户端中复制并粘贴此插件，它会毫无错误地插入记录。

我无法弄清楚问题出在哪里。任何人都可以帮助我吗？

我的代码是：

好吧，你是对的，我必须展示代码。

我在Dreamweaver中完成这个项目。实际上我读了2个数组，我创建了插入语句，我将它们插入db。

//include database file
include("dbconn.php");
//make the db connection
$myConn = new DB_Class("DB","user","password");

//initialize the string that will keep the sq   
$sqlString ="";

//outer  loop
foreach($arrayRecipients as $key2 => $row2){

    $sqlString ="INSERT INTO tbl_messages (msgTitle,msgMessage,msgDateScheduled,fkmsgAppId,msgImage,msgAction,fkmsgSerial) values (";

    //inner loop
    foreach($arrayValues as $key1 => $row1){

        $sqlString .= "'".$arrayValues[$key1]."',";
        }//end for


          //the last 3 values are fixed for now 
    $sqlString .= "'test.jpg','none','11111');</br>";

         //i call the insert function
    $myConn->insUpdRecordTest($sqlString);

        //initialize sqlstirng for the next loop
    $sqlString = "";
    }//end for

dbconn.php上的

代码是：

 CLASS DB_Class {

 VAR $db;
function DB_Class($dbname, $username, $password) {

       $this->db = mysqli_connect ('server', $username, $password)
       or DIE ("Unable to connect to Database Server");
        mysqli_select_db( $this->db,$dbname) or DIE ("Could not select database");

 }

 function query($sql) {

        $result = mysqli_query ($this->db,$sql) or DIE ("Invalid query: " . mysqli_error($this->db));
        return $result;

 }

  ///////////////////////////
 function insUpdRecordTest($sql){
     $result = $this->query($sql);

    return  $result;

 }
 }

插入语句时mysqli_query上的奇怪错误消息

0 个答案: