Inserting variable into sql string variable

Question

$sql = "insert into dbo.".$st_tab." (bike_id,st_id";
        if($st_tab === "st1_rec_id"||$st_tab === "st2_rec_id"||$st_tab ==="st3_rec_id")
        {
            $sql .= ",ln_id";
        }
        $sql .= ",times,date) values (?,?";
        if($st_tab === "st1_rec_id"||$st_tab === "st2_rec_id"||$st_tab ==="st3_rec_id")
        {
            $sql .= ",?";
        }
        $sql .= ",?,?)";
        echo $sql;
        if($st_tab === "st1_rec_id"||$st_tab === "st2_rec_id"||$st_tab ==="st3_rec_id")
        {
            $var = array($bike_id,$station,$line,$time,$date);
        }
        else
        {
            $var = array($bike_id,$station,$station,$time,$date);
        }
        $insert = sqlsrv_query($conn,$sql,$var);
        if($insert === false)
        {
            die(print_r(sqlsrv_errors(),true));
        }
        else
        {
            foreach($var as $x => $a)
            {
                echo $x." : ".$a." ";
            }
            echo "<br> 1 Record Added";
        }

due that code I got an query error

Array ( [0] => Array ( [0] => 42000 [SQLSTATE] => 42000 [1] => 102 [code] => 102 [2] => [Microsoft][ODBC Driver 11 for SQL Server][SQL Server]Incorrect syntax near 'chs_st2_rec'. [message] => [Microsoft][ODBC Driver 11 for SQL Server][SQL Server]Incorrect syntax near 'chs_st2_rec'. ) )

I can't identify what's error and where it's.

i've try to print out assembled SQL string after "$sql .= ",?,?)";" i've echo $sql but it's didn't show. thanks — Trin Pongtaman
– Trin Pongtaman, Commented Aug 7, 2015 at 9:18
Figure out what's actually going on. If you can't print out the SQL string even, then this isn't the right code excerpt. (Notably there's no chs_st2_rec to be seen; yet that's what the error message is about). If this isn't actually the relevant code, you won't get it answered. Please invest a bit more time in adding details (See How to ask, and Writing the perfect question, or MCVE…) — mario
– mario, Commented Aug 7, 2015 at 9:28
ah i've got it it's my fault the code error before this code and i get this done i've edit ** $sql = "select '$st_char' from '$st_tab' where bike_id like '$bike_id'"; ** into **$sql = "select $st_char from $st_tab where bike_id like '$bike_id'"; ** and that's work :) . thanks — Trin Pongtaman
– Trin Pongtaman, Commented Aug 7, 2015 at 9:48

Trin Pongtaman · Accepted Answer · 2015-08-07 09:53:54Z

1

figured it out

the code error before all of these line

reason is i used single quotes with variable

$sql = "select '$st_char'
            from '$st_tab'
            where bike_id like '$bike_id'";

so i remove single quotes out and it's work

thanks everyone :)

answered Aug 7, 2015 at 9:53

Trin Pongtaman

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