how to put a result of query in variable with PHP

Question

I am working with PHP and AngularJS, and I have a login.php file, Where it happens the verification of the user, and in this file, I store some information(email,token,..) in a vairable $result to use them in AngularJS. The problem that I have is that I need to store the result of a query in this variable to recuperate in AngularJS as I did for other variable. when I try this code, I recuperate the $cc variable where I stored the query result but, but it doesn't bear the query result , I get this :

cc: {current_field: "", field_count: "", lengths: "", num_rows: "", type: ""}

But in this variable I want to have the result of query. How can I do please?

login.php

<?php  

$data = json_decode(file_get_contents("php://input"));

 $connect = mysqli_connect("localhost", "root", "", "test");  

 if(count($data) > 0)  

 { 

$Email=mysqli_real_escape_string($connect, $data->Email);
$mdp=mysqli_real_escape_string($connect, $data->mdp);

$query = 'SELECT * FROM client  ';

$q = mysqli_query($connect , $query);


if(mysqli_num_rows($q) > 0 )
  {    

       $token = md5($Email.time()."51395+81519851");
       $query2 = "UPDATE client SET token = '".$token."' WHERE EmailClient = '".$Email."'";
       mysqli_query($connect , $query2);
       $_SESSION["logged_in"] = true; 
       $_SESSION["token"] = "51395+81519851"; 
       $_SESSION["Email"] = $Email; 
;
       $result['email'] =$Email;
       $result['role'] = 'client';
       $result['token'] = $token;
       $result["cc"] = $q ;


       $resultstring=json_encode($result);
       $resultstring=str_replace("null", '""', $resultstring);
       echo $resultstring;
       exit;

  }

what data is in your client table? I mean is it single record? — Haris, Aug 08 '17 at 13:46
what do you mean by **result** ? do you want the records from database? If **yes** you have to `$q=mysqli_fetch_assoc($connect,$q)` before the `if` statement. — Haris, Aug 08 '17 at 13:51
[Little Bobby](http://bobby-tables.com/) says **[your script is at risk for SQL Injection Attacks](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php)**. Learn about [Prepared Statements](http://en.wikipedia.org/wiki/Prepared_statement) for [MySQLi](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php). Even **[escaping the string](http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string)** is not safe! — GrumpyCrouton, Aug 08 '17 at 13:52
@salma If you need help learning how to use [PDO](http://php.net/manual/en/book.pdo.php) for safe and secure queries, then you can check out [this answer that I wrote](https://stackoverflow.com/a/45514591/5827005) that demonstrates a function that I wrote that makes [Prepared Statements](https://www.w3schools.com/php/php_mysql_prepared_statements.asp) **easy**, **clean**, and **secure**. Alternatively you should learn how to use [Prepared Statements for MySQLi](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) — GrumpyCrouton, Aug 08 '17 at 14:19
Thank you Anonymous, In fact it's working with mysqli_fetch, you just need to write it as response so I can mark it as right! — Salma, Aug 08 '17 at 14:37

Anandhu Nadesh · Accepted Answer · 2017-08-08T16:45:01.067

0

What you are getting is an object. The query result is an object and it is supposed to be like this. if you want to extract the rows from the object you need to use

$res=mysqli_fetch_assoc($q);

then store $res to $result['cc'] ie;

$res=mysqli_fetch_assoc($q);
$result['cc']=$res;

Note that this will display only one row from your db. If u want to display all the rows, put the fetch_assoc in a while loop. like this

$query = 'SELECT * FROM client  ';

    $s=""; 

$q = mysqli_query($connect , $query);


if(mysqli_num_rows($q) > 0 )
  {    

       $token = md5($Email.time()."51395+81519851");
       $query2 = "UPDATE client SET token = '".$token."' WHERE EmailClient = '".$Email."'";
       mysqli_query($connect , $query2);
       $_SESSION["logged_in"] = true; 
       $_SESSION["token"] = "51395+81519851"; 
       $_SESSION["Email"] = $Email; 
;
       $result['email'] =$Email;
       $result['role'] = 'client';
       $result['token'] = $token;


    while($res=mysqli_fetch_assoc($q))
    {
    $s=$s.print_r($res);
    }
       $result["cc"] = $s ;


       $resultstring=json_encode($result);
       $resultstring=str_replace("null", '""', $resultstring);
       echo $resultstring;
       exit;

  }

edited Aug 08 '17 at 16:45

answered Aug 08 '17 at 14:20

Anandhu Nadesh

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Happy to help:) – Anandhu Nadesh Aug 08 '17 at 14:37
I can't have all result, I get just one client, I don't get all client, do you have an idea please? – Salma Aug 08 '17 at 15:51
You may have to seperate it from $result[] array – Anandhu Nadesh Aug 08 '17 at 15:56
while($res=mysqli_fetch_assoc($q)){print_r( $res) } like that – Anandhu Nadesh Aug 08 '17 at 16:09
thank you but It doesn't work, I get error, and I need to put it in a variable (like $result), because I call result in javascript code! – Salma Aug 08 '17 at 16:20
It should work..I tested it. If u want to use $result then do this while($res=mysqli_fetch_assoc($q)){ $s=""; $s=$s.print_r($row); $result['cc']=$s; } – Anandhu Nadesh Aug 08 '17 at 16:34
Thank you, but when I do that I get authentication error! – Salma Aug 08 '17 at 16:41
i have modifies the answer..use it in your code. Just replace your code with the lines that i have added now – Anandhu Nadesh Aug 08 '17 at 16:45
Thank you, now I can see the difference and in console I can see every array with every client. in console everything is Ok, but it give me an authentication error that login and password are incorrect. it means something is wrong in the code – Salma Aug 08 '17 at 16:54
Ok it's working now, just we have to put $someArray[] = $res; inside the loop while, inside of print . Thank you – Salma Aug 08 '17 at 23:20

how to put a result of query in variable with PHP

1 Answers1