Mysql不会使用php插入查询

Question

This could sound stupid (maybe because it is) but I'm having so much problem in an insert query, this is the thing, I have in my first page a query that inserts into a table, the problem is that in my page2, I have EXACTLY the same query, but in this page doesn't work, the weirdest thing is that I echoed a message in the true statement of the query and it echoes, meaning that the query supposedly is already made, but is not, because when I check into the DB, is not there, and other thing! if I want to update the values, it works!, but not in create, I don't know why, but is bothering me a lot!, if someone could help me i really would appreciated, thanks.

Here is the code:

      if($radio==2){
        echo $_SESSION["in"];
        $sqlb = "SELECT * FROM table WHERE idus='$idus';";
        $resb = mysqli_query($con, $sqlb);
        $res_b = mysqli_fetch_array($resb);

        if( !$res_b)  {//if not exist on table, create
             $sqly="INSERT INTO table (x,y,z)
             VALUES ('$x','$y','$z');";  
            if ($con->query($sqly) === TRUE) {
                $_SESSION["in"]=1;
            } else { 
                echo "Error: " . $sqly . "<br>" . $con->error;
            }

        }else{ //update if exist on table
            $sqly="UPDATE table
            SET y='$y',z='$z'
            WHERE idus='$idus';";  
            if ($con->query($sqly) === TRUE) {
            } else { 
                echo "Error: " . $sqly . "<br>" . $con->error;
            }

        }

        header('Location: page2.php');
    }  

The weird thing of all is that passes through true when doing the query and do nothing, also when pasting the exact same query directly into the DB, it works.

Answer 1

It's hard to tell because we can't see all the code, but I think you should replace these lines, using as a reference the update query a few lines below:

$sqly="INSERT INTO table (x,y,z)
             VALUES (x,y,z);";  

To:

$sqly="INSERT INTO table (x,y,z)
             VALUES ('$x','$y','$z');";  

An explanation: x, y and z are not defined as SQL variables, but in another query below you use them as php variables, so I updated the code putting them as they should. Hope it helps.

Answer 2

Check your column names and properties carefully. may be x column is something wrong if update which have y,z is working fine.

Answer 3

I did not understand the question correctly but by looking at your code. You could achieve same behaviour by using INSERT...ON DUPLICATE query. see more here

if($radio==2){
        echo $_SESSION["in"];
        $sqlb = 'INSERT INTO table (x,y,z) VALUES ($x, $y, $z) ON DUPLICATE KEY UPDATE x=$x, y=$y, z=$z';
        if ($con->query($sqlb) === TRUE) {
            $_SESSION["in"]=1;
        } else { 
            echo "Error: " . $sqlb . "<br>" . $con->error;
        }

        header('Location: page2.php');
}  

Answer 4

Check the variables $x ,$y, $z variables . If its is not empty then your code works otherwise insertion don't work

Answer 5

我不知道究竟是什么问题，但我通过删除我的所有数据库结构修复它，然后重新做一遍，谢谢你的帮助！