Converting a variable to a string and more!

Question

nathmack 0 Newbie Poster

12 Years Ago

Hi, I have been trying to solve this for ages and just can’t seem to get it right.

I want to do 2 things here, i am getting an array out of a mysql database (i am getting the user’s id) and then converting that array into a string and using that string to get information out of another database.

i am trying to use the implode function $string=implode("OR id =" , $userid)
here is what i am trying to achieve: mysql_query(mysql stuf... WHERE id = $string) this should now equal: mysql_query(blah blah blah WHERE id = 1 OR id = 5 OR id = 19) because that is what is inside that vaiable now.

The problem i am having is i can't seem to get multiple rows out of my database, it only loads one.
when i echo out my array it said this Array ( [0] => 9 [ID] => 9 [1] => 19 [userid] => 19 [2] => 0 [approved] => 0 )

ALso i have tried a few differnt things with the imploe function at the moment i think it is coming up with an error.
anyway here is my code. i hope someone can help :)

<?php

$username = $_COOKIE['user'];

@mysql_connect ("localhost","down2par_down2pa","4m329aMh") or die ("could not connect to MySQL");
@mysql_select_db ("down2par_d2pdb") or die ("no database");

$query1 =mysql_query("SELECT * FROM login WHERE username = '$username'");
$user = mysql_fetch_array($query1);
$admin = $user['admin'];

if ($admin==1){

//get user id from "approval" database 
$query2 =mysql_query("SELECT * FROM approval WHERE approved = '0'");
$result =mysql_fetch_array($query2);


//turn array of ids into a string
$string = implode(" OR id =", $result['userid']);

//get total number of users
$numrow = mysql_num_rows($query2);

//use string to get users details from "login" database
$query =mysql_query("SELECT * FROM login WHERE id ='$string'");







echo ('
<html>
<head>
<style type="text/css">

td {
//padding-left: 10px;
//padding-right: 10px;
}

table {
border: solid grey 2px;
margin: 5px;
}

</style>
</head>
<h1> ads for approval: ' . $numrow . ' </h1> 
('.$string.') ');
print_r($result); 
echo('
<br><br>
');


while ($result=mysql_fetch_array($query)) {

echo ('
<table width="35%">
        <tr>
                <td width="20%"><p>user: </p>' . $result['username'] . ' </td> 
                <td width="20%"><p>club: </p>' . $result['club'] . ' </td> 
                <td width="20%"><p>city: </p>' . $result['city'] . ' </td> 
                <td width="20%"><p>night: </p>'. $result['night'] . ' </td> 
                <td width="20%"> <a href="home.php"> Approve</a> </td> 
        </tr>
</table>
<hr>
');
}
echo ('

<a href="http://www.down2party.com/Nathan/home_page/loginpage.php">Return to home</a>





</html>');
} else {echo ('you are not authorised');}
?>

mysql php

3 Contributors
6 Replies
146 Views
17 Hours Discussion Span
Latest Post 12 Years Ago Latest Post by |-|x

urtrivedi 276 Nearly a Posting Virtuoso

12 Years Ago

I have changed line 16, 20 and 26 yof your code above.
1) In 16 you are fething your row only once, you need loop through query2 result.
2) In 20 now I am using new array I set in loop, to implode.
3) In 26 I have removed single quotes around $string variable.

<?php
$username = $_COOKIE['user'];
@mysql_connect ("localhost","down2par_down2pa","4m329aMh") or die ("could not connect to MySQL");
@mysql_select_db ("down2par_d2pdb") or die ("no database");
$query1 =mysql_query("SELECT * FROM login WHERE username = '$username'");
$user = mysql_fetch_array($query1);
$admin = $user['admin'];
if ($admin==1){
//get user id from "approval" database 

$query2 =mysql_query("SELECT * FROM approval WHERE approved = '0'");
unset($arruserid);

while($result =mysql_fetch_array($query2))
{
  $arruserid[]=$result['userid'];   

}
//turn array of ids into a string
$string = implode(" OR id =", $arruserid);


get total number of users
$numrow = mysql_num_rows($query2);
//use string to get users details from "login" database
$query =mysql_query("SELECT * FROM login WHERE id =$string");






echo ('
<html>
<head>
<style type="text/css">
td {
//padding-left: 10px;
//padding-right: 10px;
}
table {
border: solid grey 2px;
margin: 5px;
}
</style>
</head>
<h1> ads for approval: ' . $numrow . ' </h1> 
('.$string.') ');
print_r($result); 
echo('
<br><br>
');
while ($result=mysql_fetch_array($query)) {
echo ('
<table width="35%">
        <tr>
                <td width="20%"><p>user: </p>' . $result['username'] . ' </td> 
                <td width="20%"><p>club: </p>' . $result['club'] . ' </td> 
                <td width="20%"><p>city: </p>' . $result['city'] . ' </td> 
                <td width="20%"><p>night: </p>'. $result['night'] . ' </td> 
                <td width="20%"> <a href="home.php"> Approve</a> </td> 
        </tr>
</table>
<hr>
');
}
echo ('
<a href="http://www.down2party.com/Nathan/home_page/loginpage.php">Return to home</a>
</html>');
} else {echo ('you are not authorised');}
?>

Edited 12 Years Ago by urtrivedi because: forgot ) in while loop,

|-|x 126 Junior Poster in Training

12 Years Ago

implode needs to be passed an array in order to work properly, what you are doing is passing a single element of an array
$result['userid'] will return the value of the userid for the first record. $result is in fact only the first record of the result set presented as an array.

Unfortunately I don't believe implode can be used to achieve what you need.
What you will need to do is loop through each record, and manually append the result value to your new query string.

for example

while ($row = mysql_fetch_array($query2)) {
  if (!empty($string))
    $string .= " OR "
  $string .= "id =".$row['userid'];
}

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nathmack 0 Newbie Poster · Answer 1 · 2012-04-16T09:06:30+00:00

perfect! thanks urtrivedi, just to clarify so i understand in the future.
you made an empty array with $arruserid[] and added into it with $result['userid'] and because you had the $results in a while it gave all values for that array one at a time, not just the first.

urtrivedi 276 Nearly a Posting Virtuoso · Answer 2 · 2012-04-16T09:22:29+00:00

$arruserid[] is used to automatically call append new useid value to array , it will give index itself from 0,1,2....n

it is similar to following code

$i=0;
 while($result =mysql_fetch_array($query2))
 {
   $arruserid[$i]=$result['userid'];   
   $i++;
 }

nathmack 0 Newbie Poster · Answer 3 · 2012-04-16T12:25:43+00:00

ok understand. thanks. Just wondering if you could help me with one more thing. i am not sure if i can do this, each time the new data is printed out in the while it is printed into a form in the form there is a submit button. i need to connect the submit button with the data in that particular form. how would i go about that?
At the moment i have a hidden field saying <input type="hidden" value="'. $result['id'] . '"> but that does not work. when i try to use that variable it is empty.

|-|x 126 Junior Poster in Training · Answer 4 · 2012-04-16T22:46:54+00:00

I think that should work, but you will need to give the hidden field a name in order to identify and grab it from the POST array.

<input type="hidden" name="id" value="'. $result['id'] . '">

then you will be able to access the data like so: $_POST['id']