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Displaying a single Database table row in php

Posted on 2011-09-04
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Last Modified: 2012-05-12
Ok, bascially, what I am trying to achieve is a piece of text that says

"Parties in CITY now"

The "CITY" is the place the user has entered in their account information and sent it to the database.

So I want to display that city row from the table on my website.

Here is my code for this,

   
mysql_connect("localhost", "username", "password") or die(mysql_error());
mysql_select_db("db_name") or die(mysql_error());

 $result = mysql_query("SELECT * FROM user");
     
    while($row = mysql_fetch_assoc($result)){
    echo " ".$row['City'].";
    } 
     

Open in new window


I get an error with  
echo " ".$row['City'].

Open in new window


So I replaced the [ ] with ( )  it gets rid of the error
but doesn't display anything on my page.

Any thoughts?
Thanks.
0
Comment
Question by:PSTCAT
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5 Comments
 
LVL 7

Expert Comment

by:dannybam
ID: 36481172
public function get_city_by_user($user) {

        $user = $this->real_escape_string($user);

        $city = $this->query("SELECT City FROM users WHERE user = '" . $user . "'");



        if ($city->num_rows > 0){

            $row = $city->fetch_row();

            return $row[0];

        } else

            return null;

    }
0
 

Author Comment

by:PSTCAT
ID: 36481180
Thanks,

but now I get

Fatal error: Using $this when not in object context


0
 
LVL 7

Accepted Solution

by:
dannybam earned 2000 total points
ID: 36481192
i took that function from a class,  it might need to be adapted...   try this

mysql_connect("localhost", "username", "password") or die(mysql_error());
mysql_select_db("db_name") or die(mysql_error());
$user = "test";
 $result = mysql_query("SELECT City FROM users WHERE user = '" . $user . "'");
       if ($result->num_rows > 0){

            $row = $result->fetch_row();

            echo $row[0];

        }
0
 

Author Comment

by:PSTCAT
ID: 36481199
Perfect. Thank you!
0
 

Author Closing Comment

by:PSTCAT
ID: 36481200
Cheers.
0

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