Foreach loop declare variable on select table array

Question

Is it possible to create variable inside foreach loop that each variable name depend on column that I select?

Here is my code:

    //sql select string declaration
    $sql = "select [Rec_ID],[Bike_ID],[Station],['Line']
            from [rfttest].[dbo].[RFT_Records_Log]
            where [Rec_ID] = '{$_GET['recid']}'";
    $query = sqlsrv_query($conn,$sql); //query
    //if query fail print out error
    if($query === false)
    {
        die(print_r(sqlsrv_errors(),true));
        sqlsrv_close($conn);
    }
    //continue with fetch array
    $recdata = sqlsrv_fetch_array( $query, SQLSRV_FETCH_ASSOC);
    //foreach to declare variable
    foreach($recdata as $x => $a)
    {
        $"$x" = $"$a";
    }

In this code I should successfully declare variable like:

$Rec_ID , $Bike_ID , $Station , $Line

I still get a syntax error:

Parse error: syntax error, unexpected '"', expecting variable (T_VARIABLE) or '$'

u cant name variable dynamically so `$"$x" = $"$a";` to `$x=$a;` thats were the parse error is from — Mystro, Aug 26 '15 at 13:28
@Mystro You can name dynamically in PHP. See http://php.net/manual/en/language.variables.variable.php. But it seems this situation is different. — Manwal, Aug 26 '15 at 13:30
@Manwal Thanks , but i meant how it was being done in the current situation Thanks tho. Lesson learnt — Mystro, Aug 26 '15 at 13:33
[The INFORMATION_SCHEMA COLUMNS Table](http://dev.mysql.com/doc/refman/5.0/en/columns-table.html) . `SELECT COLUMN_NAME, DATA_TYPE, COLUMN_DEFAULT FROM INFORMATION_SCHEMA.COLUMNS WHERE table_name = 'table_name';` — mdamia, Aug 26 '15 at 14:32

Manwal · Accepted Answer · 2015-08-26T13:34:52.220

2

Simply you can do this:

foreach($recdata as $x => $a)
{
    $x = $a;
}

Use following query:

$sql = "select [Rec_ID],[Bike_ID],[Station],['Line']
        from [rfttest].[dbo].[RFT_Records_Log]
        where [Rec_ID] = '{".$_GET['recid']."}'";

or escape ' with \ like:

$sql = "select [Rec_ID],[Bike_ID],[Station],['Line']
    from [rfttest].[dbo].[RFT_Records_Log]
    where [Rec_ID] = \'{$_GET['recid']}\'";

edited Aug 26 '15 at 13:34

answered Aug 26 '15 at 13:29

Manwal

23,450
12
63
93

You helped me once :) thanks – Trin Pongtaman Aug 26 '15 at 13:59

score 0 · Answer 2 · answered Aug 26 '15 at 13:27

0

try without quotes

$$x = $$a;

.

answered Aug 26 '15 at 13:27

Durim Jusaj

81
5

score 0 · Answer 3 · edited Aug 26 '15 at 13:36

0

I think an array is want you need.

  $data["$x"] = $a;

edited Aug 26 '15 at 13:36

Saty

22,443
7
33
51

answered Aug 26 '15 at 13:30

Jason K

1,406
1
12
15

Foreach loop declare variable on select table array

3 Answers3

Linked