problem with array

[Ameslee]

[code]
<?php
echo "<a href = \"?search=$category[0]\"> Employment </a> <br>
                      <a href = \"?search=$category[1]\"> Government </a> <br>
                      <a href = \"?search=$category[2]\"> Training </a> <br>
                      <a href = \"?search=$category[3]\"> Rehabilitation </a><br>
                      <a href = \"?search=$category[4]\"> Accommodation </a><br>
                      <a href = \"?search=$category[5]\"> Sport and Leisure </a><br>
                      <a href = \"?search=$category[6]\"> Information </a> <br>
                      <a href = \"?search=$category[7]\"> Personal Support </a> <br>
                      <a href = \"?search=$category[8]\">Education </a> <br>
                      <a href = \"?search=$category[9]\"> Transition to Work </a> <br>
                      <a href = \"?search=$category[10]\"> Community Participation </a><br>
                        </p>";
$hostname = localhost;
$username = ;
$password = ;


$conn = mysql_connect($hostname, $username, $password) or die(mysql_error());
$connection = mysql_select_db("greenac_VORNExpo", $conn);


if (in_array($_GET['search'], $category)) {

$sql = mysql_query("SELECT * FROM services WHERE category LIKE '".$search."%';") or die(mysql_error());
while ($row = mysql_fetch_array($sql)){
// display results
echo "<br><a href=\"display_services.php?id=".$row['service2id']."\">" .$row['name']. "<br>";

}

}
?>[/code]

can someone tell me why i am getting the following error and how to fix it, thanks

error:  Warning: in_array() [function.in-array]: Wrong datatype for second argument in /home/greenac/public_html/search.php on line 241

[Jessica]

is $category an array? Where is it defined?

[Ameslee]

i thought i had already done so, how do i do this?

thanks

[Jessica]

Where? You didn't do anything with it. If you mean in the links, that doesn't make any sense.
I think you want something like this

[code]$category = array('Employment', 'Government' ); // etc [/code]
then
[code]
foreach($category AS $id=>$cat){
  echo '<a href="?search='.$id.'">'.$cat.'</a> <br>';
}
[/code]
Finally
[code]if ($category[$_GET['search']]){[/code]
Try that.

[Ameslee]

[quote author=jesirose link=topic=122194.msg503697#msg503697 date=1168653569]
[code]
foreach($category AS $id=>$cat){
   echo '<a href="?search='.$id.'">'.$cat.'</a> <br>';
}
[/code]
[/quote]

Just wondering what $cat is?

[Jessica]

it's the value in the array. Foreach works like this:
$arr = array($key=>$value);

So foreach is
foreach($arr AS $key=>$value){
}

But you can call $key and $value whatever you want. $cat is short for $category.

[Ameslee]

so i have to change those too suit what i have or what u have put is ok to put straight into my code?

[Jessica]

you can put what I wrote for you.

Should search be a number or string? If it's the string, IE: Government, change this line also:
$sql = mysql_query("SELECT * FROM services WHERE category LIKE '".$category[$_GET['search']]."%';") or die(mysql_error());

[Ameslee]

is this right
[code]
<?php
$category = array('Employment', 'Government', 'Training', 'Rehabilitation', 'Accommodation', 'Sport and Leisure', 'Information', 'Personal Support', 'Education', 'Transition to Work', 'Community Participation');
foreach($category as $cat) {
  echo '<a href="?search='.$id.'">'.$cat.'</a> <br>';
}
$hostname = localhost;
$username = ;
$password = ;


$conn = mysql_connect($hostname, $username, $password) or die(mysql_error());
$connection = mysql_select_db("greenac_VORNExpo", $conn);


if ($category[$_GET['search']]){

$sql = mysql_query("SELECT * FROM services WHERE category LIKE '".$category[$_GET['search']]."%';") or die(mysql_error());
while ($row = mysql_fetch_array($sql)){
// display results
echo "<br><a href=\"display_services.php?id=".$row['service2id']."\">" .$row['name']. "<br>";

}

}
?>[/code]

[Jessica]

Looks good, try it!

[Ameslee]

hey i had a bit of trouble with that but i fixed it, one problem though, here is the code:
[code]
<?php
$category = array('Employment', 'Government', 'Training', 'Rehabilitation', 'Accommodation', 'Sport and Leisure', 'Information', 'Personal Support', 'Education', 'Transition to Work', 'Community Participation');
foreach($category as $cat) {
  echo '<a href="?search='.$cat.'">'.$cat.'</a> <br>';
}
$hostname = localhost;
$username = ;
$password = ;


$conn = mysql_connect($hostname, $username, $password) or die(mysql_error());
$connection = mysql_select_db("greenac_VORNExpo", $conn);


if (in_array($_GET['search'],$category)) {

$sql = mysql_query("SELECT * FROM services WHERE category LIKE '".$_GET['search']."%';") or die(mysql_error());
while ($row = mysql_fetch_array($sql)){
// display results
echo "<br><a href=\"display_services.php?id=".$row['service2id']."\">" .$row['name']. "<br>";

}

}
?>[/code]

ill try to explain the problem as best as possible.  I click a link, it displays a record.  if i have more than one category in that field, it wont display it, if i click a link that is a second record, it wont display.  so it doesnt look up the second thing in that field

[Ameslee]

ive worked it out thanks!

[Jessica]

Good job :)