[Help with a database query]

Mchl thanks man it was the string quotes I got it to work. I changed the combo box to a normal text box also. Thanks a lot

[Help with a database query]

yep

[Help with a database query]

Okay so i tried just hard coding the variable in the query

 

$query ="SELECT * FROM incidents WHERE inc_driverpermit='.qw.'"; //query set up

 

And got nothing but a a blank page being displayed

[Help with a database query]

sorry about the incorrect code tags.

 

It not working......I just using the an actual value in the sql statement and it says

Unknown column 'permit number here' in 'where clause'

[Help with a database query]

Hi am havign a problem where I want to return results based ona combo box where the values are generated for a table and when you make a selection it take that value and returns all results based on the value selected. here is the code.

 

combo select

<code>

 

<form action="RecordView.php" method="POST">

 

<?php

require ("dbconnect.php");

 

    $query2 = "SELECT * FROM incidents";

    $result = mysql_query($query2) or die (mysql_error());

?>

    Please select where you would like to go:<br/>

    <select name="cmbper" id="cmbper">

 

<?php

    While($row=mysql_fetch_assoc($result)){

      printf("<option value = %s> %s </option>", $row['inc_driverpermit'], $row['inc_driverpermit']);

    }//end..while

?>

<input type="Submit" name="Submit" value="GO">

  </form>

 

</code>

 

 

Record View

<code>

 

<?php

require ("dbconnect.php");

 

  // retrieve all the rows from the database

 

$cmbper=$_POST['cmbper'];

 

echo $cmbper;

 

$query ="SELECT * FROM incidents WHERE inc_driverpermit=".$cmbper.""; //query set up

  $results = mysql_query($query,$db) or die (mysql_error()); //executes query

 

  if(mysql_num_rows($results) > 0) {

  ?>

  <table border="1">

<tr>

<td>Location</td>

<td>Date</td>

<td>Police Badge</td>

<td>Vehicle Number</td>

<td>Vehicle Make</td>

<td>Vehicle Model</td>

<td>Vehicle Colour</td>

<td>Driver Frist Name</td>

<td>Driver Middle Name</td>

<td>Driver Last Name</td>

<td>Driver Permit</td>

<td>Driver Permit Expiry</td>

<td>Insurer</td>

<td>Insurance Expiry</td>

<td>Notes</td>

<td>Type</td><!--table data last column-->

</tr>

  <?php

 

  // print out the results

        while( $row = mysql_fetch_assoc( $results ) )

      {

        // print out the info

          printf("<tr>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

<td>%s</td>

</tr>", $row["inc_location"], $row["inc_datetime"], $row["inc_policebadge"], $row["inc_vehiclenum"], $row["inc_vehiclemake"], $row["inc_vehiclemodel"], $row["inc_vehiclecolour"], $row["inc_driverfirstname"], $row["inc_drivermiddlename"], $row["inc_driversurname"], $row["inc_driverpermit"], $row["inc_driverpermitexpiry"], $row["inc_insurer"], $row["inc_insuranceexpiry"], $row["inc_notes"], $row["inc_type"]);

      }

  }

?>

 

 

</code>

 

 

 

[use of undefine constant..?]

I read somewhere that php has changed it quoting rules either 'from or for' the 4.3 version.

Not sure but i think it could be something with that.

[Inserting into database]

You need to POST the email first. eg.

 

$email=$_POST['email'];

$newsletter = "INSERT INTO newsletter (email) Value ('$email')";
$rst = mysql_query($newsletter) or die("Could Not Insert");


[\code]


The last line of code the "or die" statement could be removed, I use it to know if i have a problem with my queries

[Date of birth field]

thanks the php worked thanks a million.

 

Still working on the validation for it

[Date of birth field]

is there a javascript i can use

[Date of birth field]

ok thanks i will try it

 

if i need any more help i will send the code.

 

BTW do you know any good tutorials in to validate the field so when some one choose the month the correct number of days shows up in the Day drop down menu.

[Date of birth field]

Hi Guys

 

I am have registration form with the date of birth field as a single text box but some user were not entering the format of the date of birth correctly. I decided to change from a single text bow to three drop down menu.

 

What I would like to know is it possible to get these three values to be inserted into the database as one value. eg 1982/12/30

If it is can some one help me with it.

[Need help with adding a value to a stored value in database]

Thanks Guys the code work perfectly......

[Need help with adding a value to a stored value in database]

thanks guys
am kinda new to this php and posting in forums


so i should use the [code][\code] so you will know exactly where the code starts and ends


going to try it out now thakns again[/code]

[Need help with adding a value to a stored value in database]

I am doing writing a script that updates an attribue in the database where the attribute needs to be updated by a value of 1.
eg. NoEnrolled = 12
when the script runs the number enrolled should say 13


This is the script that i am currently using but not getting it to update



<?php

  $query3 = "SELECT * FROM Class Where ClassId='".$ClassId."'";

  $result3 = mssql_query($query3, $db) or die ("Error with Query2");
         
  while($row2=mssql_fetch_array($result3)){
 
      $space = $row2['NoEnrolled'] + 1;
   
      $query5 = "UPDATE Class SET NoEnrolled =".$space." WHERE ClassId='".$ClassId."'";
      mssql_query($query5) or die ("Error with Query4");

 
  }//end..while
  ?>

[Can some help me with Microsoft SQL Server 2005 connection]

I am creating a web site using Microsoft SQL Server 2005 and using php. before i used MYSQL and php and everything was fine but now i need to know how to change the connect code from the one i use with mysql to work with Microsoft SQL Server 2005.

here is the code i used first can someone show me the modification i need to make.

<?php

$db = mysql_connect("localhost", "root","") or die ("Connection to the db died");
mysql_select_db("lis",$db) or die ("connection to lis db died");

?>