Laravel Undefined column: 7 ERROR: column "1" does not exist

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Laravel Undefined column: 7 ERROR: column "1" does not exist

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#1 by (0 votes)

3

In join , I am passing a direct value, and Laravel is trying to "read" this value as if it were a table.

function ($join) {

    $join
       ->on('e.pee_fk_pes_codigo', '=', 'p.pes_codigo')
       ->on('e.pee_padrao', '=', '1');
});

php postgresql laravel

asked by anonymous 10.08.2016 / 15:41

1 answer

Dropdown menu does not appear [closed] Performance in multi-loop applications

score 0 · Accepted Answer

Using `DB::Raw()`

The class JoinClause (used internally within Closure , which you called in join , by Laravel) by default expects value with the name of the fields of related tables.

To use a number, you can use DB::raw() . So:

function ($join) {

    $join
        ->on('e.pee_fk_pes_codigo', '=', 'p.pes_codigo')
        ->on('e.pee_padrao', '=', \DB::raw('1'));
})

DB:raw will cause Laravel to accept an SQL expression to be passed as a parameter of on .

Use of `Where`

In another form, which I believe is the best one, you can also use the where clause instead of on :

function ($join) {

    $join
        ->on('e.pee_fk_pes_codigo', '=', 'p.pes_codigo')
        ->where('e.pee_padrao', '=', '1');
})

See how the result is compiled into SQL:

$sqlString = Usuario::join('niveis', function ($j) { 

    $j->on('niveis.id', '=', 'usuarios.nivel_id')
       ->where('niveis.id', '=', 4); 

})->toSql();

 echo $sqlString;

Result:

select * from 
    'usuarios'

inner join 
    'niveis' on 'niveis'.'id' = 'usuarios'.'nivel_id' 
    and 'niveis'.'id' = ?

Laravel Undefined column: 7 ERROR: column "1" does not exist

1 answer

Using DB::Raw()

Use of Where

Using `DB::Raw()`

Use of `Where`