PHP Database output. Arrays

Question

In first case, i will get full array:

   <?php

   $servername = "localhost";
   $username = "root";
   $password = "";
   $dbname = "test.com";


   $conn = mysqli_connect($servername, $username, $password, $dbname);

   if (!$conn) {
       die("Connection failed: " . mysqli_connect_error());
   }

   $sql = "SELECT * FROM articles WHERE writer='$w_name' ";
   $result = mysqli_query($conn, $sql);


   if (mysqli_num_rows($result) > 0) {

       while ($row = mysqli_fetch_assoc($result)) {
            echo $row[header];
       }
   }
   mysqli_close($conn);

   ?>

In the second case, i will get just 1st value from array:

   <?php

   $servername = "localhost";
   $username = "root";
   $password = "";
   $dbname = "test.com";


   $conn = mysqli_connect($servername, $username, $password, $dbname);

   if (!$conn) {
       die("Connection failed: " . mysqli_connect_error());
   }

   $sql = "SELECT * FROM articles WHERE writer='$w_name' ";
   $result = mysqli_query($conn, $sql);


   if (mysqli_num_rows($result) > 0) {

       while ($row = mysqli_fetch_assoc($result)) {
            $q=$row[header];
       }
   }
   mysqli_close($conn);

   ?>

<div>
  <? echo $q ?>
</div>

What should I do, to make 2nd case work like 1st? I mean, how to put into $q, all values of array, not just the first.

Your usage of mysqli_num_rows($result) is pointless in this case. — Playdome.io
– Playdome.io, Commented Feb 5, 2015 at 14:57

davidkonrad · Accepted Answer · 2015-02-05 14:52:57Z

2

You are not defining $q as an array at all.

$q=array();
...
while ($row = mysqli_fetch_assoc($result)) {
   $q[]=$row['header'];
}
...

an example of output :

foreach($q as $header) {
   echo $header.'<br>';
}

answered Feb 5, 2015 at 14:52

davidkonrad

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Comments

satchcoder · Accepted Answer · 2015-02-05 14:49:55Z

0

You are overwriting $q . Try using $q[]=

answered Feb 5, 2015 at 14:49

satchcoder

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Legionar · Accepted Answer · 2015-02-05 14:51:11Z

0

change $q=$row[header]; line to $q[]=$row[header];

and then not echo $q; but print_r($q);

answered Feb 5, 2015 at 14:51

Legionar

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Comments

martinezjc · Accepted Answer · 2015-02-05 14:52:45Z

0

Try to change this into the while

$q = array();

if (mysqli_num_rows($result) > 0) {

       while ($row = mysqli_fetch_assoc($result)) {
            array_push($q, $row['header']); // i guess header is a column of the table
       }
}

and then you must have your array when print the $q variable, i mean do an print_r($q)

More info about array_push: http://php.net/manual/es/function.array-push.php

Hope this helps :)

answered Feb 5, 2015 at 14:52

martinezjc

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1 Comment

Playdome.io Over a year ago

if header is already defined your code won't work. Because php will not interpret it as a string.

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