The console displays only the first result of server api from mysql

Ionic Framework ionic-v3
1

hello ,
iwanted to get the result of this query to display all the results in my ionic 3 application but it returns only the first
row .
<php

$item=array();
$q = mysqli_query($con," SELECT * FROM reclamation WHERE cin_cit = ‘$postjson[cin]’ ");

while( $r=mysqli_fetch_array($q) )
{

$item = array (
‘rec_id’ => $q[‘rec_id’],
‘type_rec’ => $q[‘type_rec’],
‘description’ => $q[‘sujet’],
‘adresse’ => $q[‘adresse’],
‘laptitude’ => $q[‘laptitude’],
‘longitude’ => $q[‘longitude’],
‘etat_rec’ => $q[‘etat_reclamation’],
‘datetime’ => $q[‘datetime’],
‘gouvernorat’ => $q[‘nom_gouv’],
‘commune’ => $q[‘nom_com’],
‘cin_admin’ => $q[‘cin_admin’],
‘postal’ => $q[‘postal’],
‘fichiers’ => $q[‘justification’]
);

}

$result = json_encode(array(‘items’=>$item));
echo$result;

?>

unable to solve this ,help pls , how to put all rows in this array to send it to my ionic 3 application?

2

Do you get all rows from you server in your JSON data?

1 Like
3

Hello yes , i wanted to get all the rows in json data
here the capture of server response Capture
as u see the array do not have all rows

4

Can you just print the MySQL query string and see if query is correctly formed or not?

1 Like
5

Capture

6

here the mysql query in php code
```
$item=array();
$q = mysqli_fetch_array(mysqli_query($con, " SELECT * FROM reclamation WHERE cin_cit = ‘$postjson[cin]’ "));

      $item[] = array (
        'rec_id' => $q['rec_id'],
        'type_rec' => $q['type_rec'],
        'description' => $q['sujet'],
        'adresse' => $q['adresse'],
        'laptitude' => $q['laptitude'],
        'longitude' => $q['longitude'],
        'etat_rec' => $q['etat_reclamation'],
        'datetime' => $q['datetime'],
        'gouvernorat' => $q['nom_gouv'],
        'commune' => $q['nom_com'],
        'cin_admin' => $q['cin_admin'],
        'postal' => $q['postal'],
        'fichiers' => $q['justification']
      );

      $result = json_encode(array('items'=>$item));
      echo $result;
7

No I am saying what is the query that gets formed? Print that query and see if there is any issue with the query.

8

this is the query that i will get it’s result in ionic
$q = mysqli_fetch_array(mysqli_query($con, " SELECT * FROM reclamation WHERE cin_cit = ‘$postjson[cin]’ "));

9

Can you just

echo  "SELECT * FROM reclamation WHERE cin_cit = '$postjson[cin]'";
exit();

on your server and tell me what you get?

10

Capture

11

it displays one result while there are more then one on data base

12

Ok, does it give you more than 1 result while executing the query in phpmyadmin?

1 Like
13

yes , that’s mean this query is wrong but i can not know what to add to make it correct