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up vote 1 down vote favorite
1

I'm trying to select all the rows from my database, where 'street' is LIKE the contents from an array ($streets).

Here's what I have...

    #TEXT AREA INPUT  = $streets
    $sql = "SELECT * FROM `data` WHERE `street` LIKE '%".implode("%' OR `street` LIKE     '%",$streets)."%'";
    echo $sql;
    $result = mysqli_query($con,$sql)  or die(mysql_error());
    $totalitems1 =  mysqli_num_rows($result);
    echo $totalitems1 . "<br>";
    while($row = mysqli_fetch_array($result))
    {
    echo $row['street']. "<br>";

    }

the varible $streets is exploded from a text area input, each value on a new line.

When I process this using PHP, only the rows that are LIKE the last 'OR' are returned .. (The last line in the text area). But when I copy and paste the generated SQL into PHPMYADMIN, it returns ALL the data, as expected.

What am I missing here? Thanks.

share|improve this question
1  
Try mysqli_fetch_assoc() instead of mysqli_fetch_array() – Fred -ii- Feb 19 '14 at 23:32
1  
Sidenote: You're also mixing mysqli_query with an mysql_* function mysql_error which will result in not showing the actual error message. – Fred -ii- Feb 19 '14 at 23:46
1  
Well spotted... – user3330410 Feb 19 '14 at 23:47
up vote 0 down vote accepted

My guess would be that you aren't stripping the newline from the end of each $street entry, therefore only the last entry looks valid as it probably doesn't have a trailing newline. No doubt your query probably looks like...

SELECT * FROM `data` WHERE `street` LIKE '%foo
%' OR `street` LIKE '%bar
%' OR `street` LIKE '%baz%'

The quick fix would be to use...

implode("%' OR `street` LIKE '%", array_map(function($s) use ($con) {
    return $con->escape_string(trim($s));
}, $streets))

Ideally, you should be using a prepared statement with parameter binding.

share|improve this answer
    
SUPER! Thank you very much. I guess when you've been going all day, you look over the simple things. Thanks again Phil. – user3330410 Feb 19 '14 at 23:52

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