Re: Error Querying Database

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Sent: Wednesday, December 15, 2010 2:44 PM
Subject: Re:  Error Querying Database

>> > On Wed, 2010-12-15 at 13:42 -0500, Gary wrote:
>> >> I cant seem to get this to connect.  This is to my local testing
>> >> server,
>> >> which is on, so we need not worry that I have posted the UN/PW.
>> >>
>> >> This is a duplicate of a script I have used countless times and it
>> >> worked.
>> >> The error message is 'Error querying database.'
>> >>
>> >> Some one point out the error of my ways?
>> >>
>> >> Gary
>> >>
>> >>
>> >> <form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post">
>> >> <tr>
>> >> <td>
>> >> <label>Name of Beer</label></td><td><input name="beername" type="text"
>> >> />
>> >> </td>
>> >> </tr>
>> >> <tr>
>> >> <td>
>> >> <label>Maker of Beer</label></td><td><input name="manu" type="text" />
>> >> </td>
>> >> </tr>
>> >> <tr>
>> >> <td>
>> >> <label>Type of Beer</label></td>
>> >> <td><select name="type" size="1" id="type">
>> >>   <option>Imported</option>
>> >>   <option>Domestic</option>
>> >>   <option>Craft</option>
>> >>   <option>Light</option>
>> >> </select>
>> >> <!--<select name="avail" size="1" id="avail">
>> >>   <option>Available</option>
>> >>   <option>Sold</option>
>> >> </select>-->
>> >> </td>
>> >> </tr>
>> >> <tr>
>> >> <td><label>Sold in</label>
>> >> </td><td><input type="checkbox" name="singles" value="Yes" />
>> >> Singles<br
>> >> />
>> >> <input type="checkbox" name="six" value="Yes" /> Six Packs <br />
>> >> <input type="checkbox" name="can" value="Yes" /> Cans<br />
>> >> <input type="checkbox" name="bottles" value="Yes" /> Bottles <br />
>> >> <input type="checkbox" name="tap" value="Yes" /> Draft <br />
>> >> <tr>
>> >> <td>
>> >> <label>Size</label></td><td><input name="size" type="text" />
>> >> </td></tr>
>> >> <tr><td>
>> >> <label>Description</label></td><td><textarea name="desc" cols="40"
>> >> rows="5"></textarea>
>> >> </td></tr>
>> >> <tr><td>
>> >> <input name="submit" type="submit" value="Submit" /></td></tr>
>> >> </form>
>> >> </table>
>> >> </div>
>> >> <div id="list">
>> >> <?php
>> >> $beername = $_POST['beername'];
>> >> $manu = $_POST['manu'];
>> >> $type = $_POST['type'];
>> >> $singles = $_POST['singles'];
>> >> $six = $_POST['six'];
>> >> $can = $_POST['can'];
>> >> $bottles = $_POST['bottles'];
>> >> $tap = $_POST['tap'];
>> >> $size = $_POST['size'];
>> >> $desc = $_POST['desc'];
>> >> $ip= $_SERVER['REMOTE_ADDR'];
>> >>
>> >> $dbc = mysqli_connect('localhost','root','','rr')or die('Error
>> >> connecting
>> >> with MySQL Database');
>> >>
>> >> $query = "INSERT INTO beer (beername, manu, type, singles, six, can,
>> >> bottles, tap, size, desc, ip )"." VALUES ('$beername', '$manu',
>> >> '$type',
>> >> '$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc',
>> >> '$ip' )";
>> >>
>> >> $result = mysqli_query($dbc, $query)
>> >> or die('Error querying database.');
>> >>
>> >>
>> >> mysqli_close($dbc);
>> >>
>> >>
>> >>
>> >> -- 
>> >> Gary
>> >
>> >
>> > Read Ash's reply...   but basically, you're running the query with POST
>> > variables, and inserting them on page display as well as on form
>> > submit.
>> >
>> > can you ensure that you can connect from the command line?
>> >
>> >
>> > if you may take some criticism, you should rethink your database
>> > design,
>> > as well as the page flow/design... you should either post the form to a
>> > new page, or if it is back to itself, you should check to see that you
>> > have in fact posted it before just blindly inserting into the database
>> > (as currently, every time you view the page, you will insert into the
>> > database, even if completely empty values).
>> >
>>
>> Steve
>>
>> Thank you for your reply.
>>
>> I did not see a reply from Ashley, but I would love to read it.
>>
>> I always welcome criticism, however this form is for the owner of a bar
>> where he will inputing his list of beer that he sells.  The rest of the
>> code
>> that is not there is I will have the list then echo to screen below the
>> form.  This is an internal list only, no customers will be seeing
>> it....if
>> that makes any difference to your suggestion.
>>
>> On your one point
>>
>> <<(as currently, every time you view the page, you will insert into the
>> database, even if completely empty values).>>
>>
>> Is this always the case when you process a form onto itself?  Or is there
>> a
>> fix?
>>
>> I did just create a new page, inserted the script onto it, and got the
>> same
>> error message.
>>
>> Again, thank you for your help.
>>
>> Gary
>
>
> Gary
>
> the line:
> <input name="submit" type="submit" value="Submit" />
>
> is the submit part... if you encapsulate the DB part of the code,
> within:
> if($_POST['submit'] == 'Submit')
> {
>    # do db stuff in here and value sanitizing
> }
>
> basically, what that does is the submit button is name $_POST['submit']
> and has the value "Submit" (the 'value' of the button) which will not be
> set, until you submit the form.   Personally, I do it another way, but
> but is how most people check to see if a form is submitted (i think?)
>
> but it seems as if your issue stems from the lack of being able to
> connect to the database itself, so either your login credentials are
> wrong, or you dont have the mysqli connector enabled with your php in
> your development box.
>
> check your phpinfo() to ensure it is enabled.
>
> Steve
>
Steve

Again thank you, here is my phpini information, so it would seem mysqli is
enabled


      MysqlI Support enabled
      Client API library version  5.1.41
      Active Persistent Links  0
      Inactive Persistent Links  0
      Active Links  13
      Client API header version  5.1.41
      MYSQLI_SOCKET  MySQL

      Directive Local Value Master Value
      mysqli.allow_local_infile On On
      mysqli.allow_persistent On On
      mysqli.default_host no value no value
      mysqli.default_port 3306 3306
      mysqli.default_pw no value no value
      mysqli.default_socket MySQL MySQL
      mysqli.default_user no value no value
      mysqli.max_links Unlimited Unlimited
      mysqli.max_persistent Unlimited Unlimited
      mysqli.reconnect Off Off




So you are saying that if I (preferably on a form that is working), this
will stop the double entries?

 if($_POST['submit'] == 'Submit')
 {
    $dbc = mysqli_connect('localhost','root','','rr')or die('Error
connecting with MySQL Database MySQL said: '.mysql_error());

$query = "INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip )VALUES ('$beername', '$manu', '$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc', '$ip' )";

$result = mysqli_query($dbc, $query)
or die('Error querying database.  MySQL said: '.mysql_error());
 }

Thank you

Gary

PS. Steve, sorry, but it appears I hit the reply instead of reply to group, 
I was not intending to send you an email direct..

Gary


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__________ Information from ESET Smart Security, version of virus signature database 5706 (20101215) __________

The message was checked by ESET Smart Security.

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