CODEWITHSUNDEEP
phpmysqlarrays
BRG
BRG

Reputation: 380

Array prints data correctly but query shows only half

I am having a very very strange issue here. I am having data inside of an array in php. I want for each item that the array has to query the database and get results back. It only shows me specific results and not everything.

My code:

foreach($my as $k=> $v){
    //echo "Key: ". $k . " Value: " . $v . "<br/>";
    $sql2 = "SELECT column10 FROM `table` WHERE column1 = '$v' ";
    $res2 = mysql_command($sql2);
    echo $sql2 . "<br/>";
    $rowA = mysql_fetch_assoc($res2);
    //echo "<strong>Alternative: </strong>" . $v. "<strong> Auto Alternative: </strong>" . $rowA['column10'] . "<br/>";

}
echo '</table>';
echo "<pre>";
print_r($my);
echo "</pre>";

In the browser the result if i echo the query and Key: and value is this:

SELECT column10 FROM `table` WHERE column1 = 'Villetta La Canoa' 
SELECT column10 FROM `table` WHERE column1 = ' Casa Immerso nel Verde' 
SELECT column10 FROM `table` WHERE column1 = ' La Rosetta' 
SELECT column10 FROM `table` WHERE column1 = 'Agriturismo La Nonna' 
SELECT column10 FROM `table` WHERE column1 = ' Villetta Cassiopeia' 
SELECT column10 FROM `table` WHERE column1 = ' La Rosetta' 
SELECT column10 FROM `table` WHERE column1 = 'Ca Gianca 2' 
SELECT column10 FROM `table` WHERE column1 = ' Villetta Teresa' 
SELECT column10 FROM `table` WHERE column1 = ' Appartamento Pinamare' 
SELECT column10 FROM `table` WHERE column1 = ' Casa del Principe' 
SELECT column10 FROM `table` WHERE column1 = 'Ca Gianca 2' 
SELECT column10 FROM `table` WHERE column1 = ' Villetta Teresa' 
SELECT column10 FROM `table` WHERE column1 = ' Appartamento Pinamare' 
SELECT column10 FROM `table` WHERE column1 = ' Casa del Principe' 
SELECT column10 FROM `table` WHERE column1 = 'Ca Gianca 2' 
SELECT column10 FROM `table` WHERE column1 = ' Villetta Teresa' 
SELECT column10 FROM `table` WHERE column1 = ' Appartamento Pinamare' 
SELECT column10 FROM `table` WHERE column1 = ' Casa del Principe'

The Key,Value is :

Key: 0 Value: Villetta La Canoa
Key: 1 Value: Casa Immerso nel Verde
Key: 2 Value: La Rosetta
Key: 3 Value: Agriturismo La Nonna
Key: 4 Value: Villetta Cassiopeia
Key: 5 Value: La Rosetta
Key: 6 Value: Ca Gianca 2
Key: 7 Value: Villetta Teresa
Key: 8 Value: Appartamento Pinamare
Key: 9 Value: Casa del Principe
Key: 10 Value: Ca Gianca 2
Key: 11 Value: Villetta Teresa
Key: 12 Value: Appartamento Pinamare
Key: 13 Value: Casa del Principe
Key: 14 Value: Ca Gianca 2
Key: 15 Value: Villetta Teresa
Key: 16 Value: Appartamento Pinamare
Key: 17 Value: Casa del Principe

and what i get is:

Alternative: Villetta La Canoa Auto Alternative: Villa Ronchi, Casa Ciserai, Villino Torretta, Casa Bianca
Alternative: Casa Immerso nel Verde Auto Alternative: 
Alternative: La Rosetta Auto Alternative: 
Alternative: Agriturismo La Nonna Auto Alternative: Agriturismo Antico Granaio, Casa Ciserai, Villa Ronchi, La Rosetta
Alternative: Villetta Cassiopeia Auto Alternative: 
Alternative: La Rosetta Auto Alternative: 
Alternative: Ca Gianca 2 Auto Alternative: Ca Gianca 1, La Vigna 2, Villetta Teresa
Alternative: Villetta Teresa Auto Alternative: 
Alternative: Appartamento Pinamare Auto Alternative: 
Alternative: Casa del Principe Auto Alternative: 
Alternative: Ca Gianca 2 Auto Alternative: Ca Gianca 1, La Vigna 2, Villetta Teresa
Alternative: Villetta Teresa Auto Alternative: 
Alternative: Appartamento Pinamare Auto Alternative: 
Alternative: Casa del Principe Auto Alternative: 
Alternative: Ca Gianca 2 Auto Alternative: Ca Gianca 1, La Vigna 2, Villetta Teresa
Alternative: Villetta Teresa Auto Alternative: 
Alternative: Appartamento Pinamare Auto Alternative: 
Alternative: Casa del Principe Auto Alternative:

Can anyone explain what is wrong?

Upvotes: 2

Views: 127

Answers (2)

besciualex
besciualex

Reputation: 1892

From my point of view you may have to trim $v to be sure there are no white spaces at the start and end of string.

foreach($my as $k=> $v){
   $v = trim($v);

Also you should try the 'LIKE' operator instead of '='. Perhaps add '%' to $v.

$sql2 = "SELECT column10 FROM `table` WHERE column1 LIKE '%{$v}%'";

Upvotes: 4

Hanky Panky
Hanky Panky

Reputation: 46900

What on earth is mysql_command? I'm sure its a typo and you meant mysql_query (Alas even for that :-/)

And you dont get results for some because of an extra space behind some strings like this one

SELECT column10 FROM `table` WHERE column1 = ' Casa Immerso nel Verde' 
                                              ^

Your sample data confirms this. The only results you have no alternatives for are the ones that have that extra space.

Upvotes: 3

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