php error Catchable fatal error when making query

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php error Catchable fatal error when making query

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1

I have this doubt, it is probably very simple answer but I have very little doing this because I still do not soak up the logic hehehehe. Good to the question:

This part of the code marks me the error: "Catchable fatal error: Object of class mysqli_result could not be converted to string"

 $queryjoin="SELECT matriculas.matricula from matriculas inner join aspirantes on matriculas.id_mat=aspirantes.id_asp where aspirantes.folio=$id;";
                     $n_control = mysqli_query($link,$queryjoin); echo $n_control;

can you help me? they are two tables in a same base, when I run it in the consulate of mysql if it gives me the desired result but here when doing the echo $ n_control; Mark me that error, help for fa in the tables the values are INT

A thousand thanks

mysql php

asked by Erick Castillo 06.06.2018 в 21:03

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1 answer

Insert correlative to an ID field in SQL Open Form of another project C # (mvc)

score 0 · Accepted Answer

The variable $ n_control contains an object of type mysqli_result. You can only use echo to display strings. The usual way is done:

$queryjoin="SELECT matriculas.matricula from matriculas inner join aspirantes on matriculas.id_mat=aspirantes.id_asp where aspirantes.folio=$id;";
$n_control = mysqli_query($link,$queryjoin);

while($row = mysqli_fetch_assoc($n_control)) {
    echo $row['matricula'];
}