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Can anyone help, why doesn't this work ?? - PHP Development

I have a from with 2 fields: Company & Name Depening which is completed, one of the following queries will be run: if($Company){ $query = "Select C* From tblsample Where ID = $Company Order By Company ASC"; } if($Name){ $query = "SELECT * FROM TBLsample WHERE Contact = "%$Name%" ORDER BY Contact ASC"; } $result = mysql_query($query); $number = mysql_numrows($result); for ($i=0; $i<$number; $i++) { $CompanyName = mysql_result($result,$i,"Company"); $ContactName = mysql_result($result,$i,"Contact"); Print "Company: $CompanyName"; Print "<P>Name: $ContactName"; } This work fine for 'Company', but not for 'Name' How do I do a wild search using $Name in the Where statement.. ...

  1. #1

    Default Can anyone help, why doesn't this work ??

    I have a from with 2 fields:
    Company & Name

    Depening which is completed, one of the following queries will be run:

    if($Company){
    $query = "Select C*
    From tblsample
    Where ID = $Company
    Order By Company ASC";
    }

    if($Name){
    $query = "SELECT *
    FROM TBLsample
    WHERE Contact = "%$Name%"
    ORDER BY Contact ASC";
    }


    $result = mysql_query($query);
    $number = mysql_numrows($result);
    for ($i=0; $i<$number; $i++) {

    $CompanyName = mysql_result($result,$i,"Company");
    $ContactName = mysql_result($result,$i,"Contact");

    Print "Company: $CompanyName";
    Print "<P>Name: $ContactName";
    }

    This work fine for 'Company', but not for 'Name'

    How do I do a wild search using $Name in the Where statement.. ?

    Ie $Name = Jim
    So %Jim% should result in :
    Jim
    jimmy
    jimm
    etc !!

    Thanks
    James Guest

  2. #2

    Default Re: Can anyone help, why doesn't this work ??

    James <Jamesnothere.com> wrote:
    > $q="SELECT ... Contact = "%$Name%" ORDER BY Contact ASC";
    ^begin string ^endstring
    ^beginstring ^endstring

    Learn the differences between " and ' and the use of these within a
    string ([url]http://nl.php.net/manual/en/language.types.string.php[/url]).

    Other problems are realted to sql.

    -wildcards (%) only work on strings AFAIK
    -strings need to be quoted (eg ').

    --

    Daniel Tryba

    Daniel Tryba Guest

  3. #3

    Default Re: Can anyone help, why doesn't this work ??

    On Sat, 28 Jun 2003 12:22:02 +0000 (UTC), Daniel Tryba
    <news_comp.lang.phpcanopus.nl> wrote:
    >James <Jamesnothere.com> wrote:
    >> $q="SELECT ... Contact = "%$Name%" ORDER BY Contact ASC";
    > ^begin string ^endstring
    > ^beginstring ^endstring
    >
    >Learn the differences between " and ' and the use of these within a
    >string ([url]http://nl.php.net/manual/en/language.types.string.php[/url]).
    >
    >Other problems are realted to sql.
    >
    >-wildcards (%) only work on strings AFAIK
    >-strings need to be quoted (eg ').
    And wildcards in SQL only apply if you use 'LIKE' instead of '='.

    --
    Andy Hassall (andyandyh.co.uk) icq(5747695) ([url]http://www.andyh.co.uk[/url])
    Space: disk usage ysis tool ([url]http://www.andyhsoftware.co.uk/space[/url])
    Andy Hassall Guest

  4. #4

    Default Re: Can anyone help, why doesn't this work ??

    My form and results are on one page.
    If I use :

    if ($Company) {
    $query = "Select Company, Contact
    From tblworking
    Where ID = $Company
    Order By Company ASC";
    }

    if ($Name ) {
    $query = "SELECT *
    FROM TBLWorking
    WHERE Contact Like '%$Name%'
    ORDER BY Contact";
    }


    $result = mysql_query($query);
    $number = mysql_numrows($result);
    for ($i=0; $i<$number; $i++) {

    $CompanyName = mysql_result($result,$i,"Company");
    $ContactName = mysql_result($result,$i,"Contact");

    Print "Company: $CompanyName";
    Print "<P>Name: $ContactName";
    }

    As soon as I enter the page it searchs using $Name, even though I have not
    entered any information.

    This results in me getting a complete list printed on screen, with out me
    having to search.

    How do I get this to show just the form, then the form and the results ?

    Thanks



    On Sat, 28 Jun 2003 15:38:53 +0100, James <JamesNotHere.com> wrote:
    >Thanks, its now working using:
    >
    >WHERE Contact Like '%$Name%'
    >
    >Cheers
    >
    >On Sat, 28 Jun 2003 13:36:19 +0100, Andy Hassall <andyandyh.co.uk> wrote:
    >
    >>On Sat, 28 Jun 2003 12:22:02 +0000 (UTC), Daniel Tryba
    >><news_comp.lang.phpcanopus.nl> wrote:
    >>
    >>>James <Jamesnothere.com> wrote:
    >>>> $q="SELECT ... Contact = "%$Name%" ORDER BY Contact ASC";
    >>> ^begin string ^endstring
    >>> ^beginstring ^endstring
    >>>
    >>>Learn the differences between " and ' and the use of these within a
    >>>string ([url]http://nl.php.net/manual/en/language.types.string.php[/url]).
    >>>
    >>>Other problems are realted to sql.
    >>>
    >>>-wildcards (%) only work on strings AFAIK
    >>>-strings need to be quoted (eg ').
    >>
    >> And wildcards in SQL only apply if you use 'LIKE' instead of '='.
    James Guest

  5. #5

    Default Re: Can anyone help, why doesn't this work ??


    "James" <JamesNotHere.com> wrote in message
    news:dh1rfvkjstndhgsmdl16g2njhgetj71r8v4ax.com...
    > I have a from with 2 fields:
    > Company & Name
    >
    > Depening which is completed, one of the following queries will be run:
    >
    > if($Company){
    > $query = "Select C*
    > From tblsample
    > Where ID = $Company
    > Order By Company ASC";
    > }
    >
    > if($Name){
    > $query = "SELECT *
    > FROM TBLsample
    > WHERE Contact = "%$Name%"
    > ORDER BY Contact ASC";
    > }
    >
    >
    > $result = mysql_query($query);
    > $number = mysql_numrows($result);
    > for ($i=0; $i<$number; $i++) {
    >
    > $CompanyName = mysql_result($result,$i,"Company");
    > $ContactName = mysql_result($result,$i,"Contact");
    >
    > Print "Company: $CompanyName";
    > Print "<P>Name: $ContactName";
    > }
    >
    > This work fine for 'Company', but not for 'Name'
    >
    > How do I do a wild search using $Name in the Where statement.. ?
    >
    > Ie $Name = Jim
    > So %Jim% should result in :if($Name){
    > Jim
    > jimmy
    > jimm
    > etc !!
    >
    > Thanks
    if ( $Company )
    {
    $query = "Select C* From tblsample Where ID LIKE '%$Company%' Order By
    Company ASC";
    }
    if ( $Name )
    {
    $query = "SELECT * FROM TBLsample WHERE Contact LIKE '%$Name%' ORDER BY
    Contact ASC";
    }


    Roger Guest

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