选择工作查询，但不插入查询

Question

  <?php
    $sql = "SELECT MAX(REQUEST_ID) AS REQUEST_ID FROM TBL_REQUEST;";
    $stmt = sqlsrv_query( $conn, $sql );
    if( $stmt === false) {
        echo "alert('DataBase Error');";
    }
    else{
        $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC);
        $newReqId=$row['REQUEST_ID']+1;
    }
    $sql = "INSERT INTO TBL_REQUEST VALUES(".$newReqId.",\'ISSUE\',\'Test\',\'Test\',\'NSNS05\',1);";
    $newsql = $sql;
    $stmt = sqlsrv_query( $conn, $newsql );
    if( $stmt === false) {
        echo "alert('Not Inserted '+assoc_id+' ".$stmt." '+' ".$newsql." ');";
    }
    else{
        echo "alert('Inserted '+assoc_id+' ".$stmt."');";
    }

?>

在上面的代码中，我使用了2条SQL语句，第一个是select语句，它可以正常工作并从表中获取数据，但是当我在同一张表上执行Insert时，它将无法工作...

INSERT INTO TBL_REQUEST VALUES(4016,'ISSUE','Test','Test','NSNS05',1); 

多数民众赞成在字符串是

Answer 1

为什么使用""间""你应该' '为$newReqId

$sql = "INSERT INTO TBL_REQUEST VALUES(".$newReqId.",\'ISSUE\',\'Test\',\'Test\',\'NSNS05\',1);";