AJAX AND PHP

[etsted]

$type = $_REQUEST['type'];        $sql = "SELECT tittel FROM $type GROUP BY titel";    $query = mysqli_query($con,$sql);    while($row = mysql_fetch_array($query))    {        $a[] = $row['titel'];    }        // get the q parameter from URL    $q=$_REQUEST["q"]; $hint="";        // lookup all hints from array if $q is different from ""     if ($q !== "")      { $q=strtolower($q); $len=strlen($q);        foreach($a as $name)        { if (stristr($q, substr($name,0,$len)))          { if ($hint==="")            { $hint=$name; }            else            { $hint .= "<br> $name"; }          }        }      }        // Output "no suggestion" if no hint were found    // or output the correct values     echo $hint==="" ? "no suggestions" : $hint;exit();

this script is supposed to show suggested titels according to what the user is typing in.

 

But if i have several titel in the DB like this:

"my new name"

"name"

"new"

and i were to search "my new name" then all of the above titels will show, when only the first should.

[davej]

Where do you define the $con variable?

[etsted]

why does that matter?

its actually right above where i call it like include_once("connect.php");

[dsonesuk]

That is because you are searching through 'haystack' ($q) for 'needle' of "my new name" or "name" or "new", which it will find 'my new name' and pass 'if' condition and add to $hint as $hint is empty, loop continues $hint not empty now so

go through 'else' condition $hint .= "<br> $name"; until end of loop.

 

You can either swap $q and $name around, or

if (strripos($q, $name) != -1)

or

if ($name == $q)

depending on what you require

[etsted]

ive tried to update it, and it almost work as i want i to but there are some small problems.

    if ($q !== "")      { $q=strtolower($q); $len=strlen($q);        foreach($a as $name)        {            if (strcmp(substr($name,$len), substr($q,0,$len)))          {               if ($hint==="")                { $hint=$name; }              else                { $hint .= "<br> $name"; }          }        }      }   

this script will shows you only what you search for, but if you search for "n"

and i have:

"green"

"nice tv"

then both of them will show, basically i want to make sure that when someone search for something the only matches should have what the user searched for at the beginning of the name, Not in the middle or the end or something like that.

[etsted]

shoudnt this work?

if ($q !== "")

{ $q=strtolower($q); $len=strlen($q);

foreach($a as $name)

{

if ( (strcmp(substr($name,0,$len), substr($q,0,$len)) ) == true)

{

if ($hint==="")

{ $hint=$name; }

else

{ $hint .= "<br> $name"; }

}

}

}

Because it dont

 

In the DB i have 4 titels "6", "d", "green" and "nature" and if i type "n" then 6,d and green shows up then i try to type in "nature " (notice the space i put after nature) and then nature shows up at the end

Edited September 8, 2014 by etsted

[etsted]

i noticed something else, it doesnt show titel that does have a match.

[dsonesuk]

Well you could filter directly from sql query, to show title that start with, then filter it further.

SELECT tittel FROM $type WHERE title LIKE '$q%' GROUP BY titel

[etsted]

why dont i put the % at the front?

[dsonesuk]

then both of them will show, basically i want to make sure that when someone search for something the only matches should have what the user searched for at the beginning of the name, Not in the middle or the end or something like that.

 

 

because i thought that is what you wanted by this quote in red, if you place at beginning as well it will find your search criteria anywhere within text string.