Help on a db query please.

Discussion in 'Databases' started by JEET, Jul 10, 2006.

0
  1. #1
    Hi,
    I am getting a value submitted from a form.

    $val= $_POST['field'];
    Then putting it in an array.
    $arr= explode("/", $val);

    Now I have to open a db and check if this value in that field already exists or not. if not, insert it.
    This is what I am not getting.

    mysql_connect(localhost,username,password);
    @mysql_select_db($database) or die( "Unable to select database");
    $query="SELECT * FROM $tablename WHERE field='$cat'";
    $result=mysql_query($query);
    $num=mysql_numrows($result);
    mysql_close();

    I know that I need to run a foreach loop to automatically check for each item in $arr but I am completely confused by the number of records in $num now.

    My question:
    What code needs to be put in FOREACH loop now.
    It has to check all records selected for the previous query with the current value in $arr.
    If a record with $arr value is found, do nothing or else insert a record.

    foreach ($arr as $value)
    {

    }

    Thank you for any help here.
     
    JEET, Jul 10, 2006 IP
  2. T0PS3O

    T0PS3O Feel Good PLC

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    #2 
    
foreach ($arr as $value)
{
$query="SELECT * FROM $tablename WHERE field='" . $value . "'";
$result=mysql_query($query);
$num=mysql_numrows($result);
if ($num == 0) {
//Insert
}
}
    PHP:
     
    T0PS3O, Jul 10, 2006 IP
  3. JEET

    JEET Notable Member

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    #3
    Hi,
    Thanks so much.
    It was easy just had to let go the pressure.
    Thanks again.
    Here's what I came up with. ( a little longer than what you gave but doesn't involve querying the db over and over)
    ----------------
    Code in first post here.

    foreach ($arr as $value)
    {

    $i=0;
    $c= "1";
    while ($i < $num) {
    $f = mysql_result($result,$i,"field");
    if ($f == $value)
    $c= "2";

    ++$i;
    }

    if ($c== "1")
    {
    INSERT here
    }
    }
    -----------------
    Bye
     
    JEET, Jul 10, 2006 IP