[RESOLVED] Array only giving first letter! :(

thegoldenchild

Hi,

EDIT: Im trying to get info from a database, i managed to i think, but the only thing i get from the array is the first letter 🙁.

$connection = mysql_connect($host, $dblogin, $dbpassword) or die(mysql_error("connection klopt niet")); 
$db = mysql_select_db($dbname, $connection) or die(mysql_error(("db klopt niet"))); 

$query = "SELECT * FROM aanmelding WHERE woonplaats='$_POST[search]'"; 
$result = mysql_query($query) or die(mysql_error()."werkt niet 2");
$row = mysql_fetch_array($result, MYSQL_ASSOC);

if ($row == "")
{
echo ($search ." komt niet in de database voor, misschien is er nog niemand anders die zich heeft
 aangemeld vanuit jou woonplaats. Kijk of je een typefout hebt gemaakt
 of probeer een andere woonplaats.<br><br> Meld jezelf wel aan zodat iemand anders misschien dit probleem
  niet meer heeft!");
}
else
{

echo $row[naam][0];


}
}
?>

rincewind456

Get rid of the

[0]

on the end of the line

echo $row[naam][0];

and see what happens

Rodney_H_

I notice a couple of things with your code:

1) I believe rincewind456 is correct

2) $POST[search] and $row[naam] need quotes for your array keys: $POST['search'] $row['naam']

3) You are using the poste value directly in your SQL query without processing it, sanitizing it against user input that may be malicious, which could cause an sql injection attack (or whatever the kids are calling it these days).

I highly recommend that you add slashes if needed, and process the posted value BEFORE using it to query the database with for the sake or your site's security.

thegoldenchild

Thanx for your replies.

I've removed the [0] with the result that the first name is given. But it doesn't come up with the others.

I don't really know how to "protect" input from users, i'm only using strip_tags atm.. Rincewind, did show me another, but i don't the format yet..i'm going to check it out now..

Is it possible to put the array in a foreach?

Since i don't completely understand the foreach function.. :o

foreach ($naam as ? => ?)
{
And this would be repeated untill all rows are done..or something like that
}

But i don't know what i should enter between ().

EDIT: I found out that "mysql_real_escape_string()" is just as easy as strip_tags, so im using it now 🙂 wont forget it anymore..

EDIT: Some more information:

The table has the following fields/rows:

naam (Name1, Name2, Name3)
woonplaats (City1, City2, City3)
leeftijd (Age1, Age2, Age3)
email (Email1, Email2, Email3)

So there are more things then one if ur search query isn't just based on 1 "person/city".

It should become something like. That he loads all info from db that it needs. After that it becomes in a function counting untill it hits the nr inside de db of rows. It will leap untill it had all rows echo-ing but within 1 function.


$nr1 = "0";

function show($row)
{
count nr of rows of table = $becomes variable;

max. repeating for all rows = $becomes variable;


$nr1 ++ 1;


if ( $becomes variable == $nr1)
{
break
}
}

I don't know if this is allright, but at least i tried lol

rincewind456

Use a while loop instead of a foreach like

while ($row = mysql_fetch_row($result)){
	echo $row['naam'];
}

thegoldenchild

After trying your answer:

$connection = mysql_connect($host, $dblogin, $dbpassword) or die(mysql_error("connection klopt niet")); 
$db = mysql_select_db($dbname, $connection) or die(mysql_error(("db klopt niet"))); 

$query = "SELECT * FROM aanmelding WHERE woonplaats='$_POST[search]'"; 
$result = mysql_query($query) or die(mysql_error()."werkt niet 2");
$row = mysql_fetch_array($result, MYSQL_BOTH);

if ($row == "")
{
echo ($search ." komt niet in de database voor, misschien is er nog niemand anders die zich heeft
 aangemeld vanuit jou woonplaats. Kijk of je een typefout hebt gemaakt
 of probeer een andere woonplaats.<br><br> Meld jezelf wel aan zodat iemand anders misschien dit probleem
  niet meer heeft!");
}
else
{

while ($row = mysql_fetch_row($result))
{ 
    echo $row['naam']; 
} 

}
}
?>

I don't get anything returned from the query :S 🙁

EDIT:

I worked it out


//mysql_fetch_row should have been mysql_fetch_array

while ($row = mysql_fetch_row($result))
{ 
    echo $row['naam']; 
}

But after that i don't get the first row from the table ( nr 0 in sql) i do get the second..and 3rd en so on..

thegoldenchild

i can fill up the first one with a blank sheet, so it doesn't miss anything. But if u ask me that's not the way.. 😕

thegoldenchild

$query = "SELECT * FROM aanmelding WHERE woonplaats='$_POST[search]'"; 
$result = mysql_query($query) or die(mysql_error()."werkt niet 2");
$row = mysql_fetch_array($result, MYSQL_BOTH);

if ($row == "")
{
echo ($search ." komt niet in de database voor, misschien is er nog niemand anders die zich heeft
 aangemeld vanuit jou woonplaats. Kijk of je een typefout hebt gemaakt
 of probeer een andere woonplaats.<br><br> Meld jezelf wel aan zodat iemand anders misschien dit probleem
  niet meer heeft!");
}
else
{
echo("<div align='center'><table border='0' cellspacing='0' cellpadding='0' bordercolor=''><tr><td width='90'><font size='2' face='arial' color='#FF6000'><b>Naam</b</font></td><td width='100'><font size='2' face='arial' color='#FF6000'><b>Woonplaats</font></td><td width='80'><font size='2' face='arial' color='#FF6000'><b>Leeftijd</font></td>
<td width='80'><font size='2' face='arial' color='#FF6000'><b>Firmware</font></td><td width='250'><font size='2' face='arial' color='#FF6000'><b>Email</font></td></tr>");

echo ("<tr>
    <td><font size='2' face='arial' color='#000000'>" . $row['naam'] . "</font></td>
    <td><font size='2' face='arial' color='#000000'><b>" . $row['woonplaats'] . "</b></font></td>
    <td><font size='2' face='arial' color='#000000'>" . $row['leeftijd'] . "</font></td>
    <td><font size='2' face='arial' color='#000000'>" . $row['firmware'] . "</font></td>
    <td><font size='2' face='arial' color='#000000'>" . $row['email'] . "</font></td>
    </tr>");

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

echo ("<tr>
<td><font size='2' face='arial' color='#000000'>" . $row['naam'] . "</font></td>
<td><font size='2' face='arial' color='#000000'><b>" . $row['woonplaats'] . "</b></font></td>
<td><font size='2' face='arial' color='#000000'>" . $row['leeftijd'] . "</font></td>
<td><font size='2' face='arial' color='#000000'>" . $row['firmware'] . "</font></td>
<td><font size='2' face='arial' color='#000000'>" . $row['email'] . "</font></td>
</tr>"); 

} 
echo("</table></div>");
}
}

Fixed it myself 🙂 took some time..and it's probably not the most obvious way..but it works 🙂