UpdateDB in Action - MySQL

Stop writing insert and update SQL statements and cut the time you spend writing simple SQL in half while focusing on the more complicated things. Leave it up to OOP to help you out. We will make a class that goes out and looks for the values for us and builds a SQL statement on the fly. All we have to do is make sure the column names in the database correspond with the field names in the HTML form. Believe me when I say it saves TONS of time. I never write applications that don't use it.

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By: Sam 'SammyK' Powers

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February 04, 2004

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This will be similar to how we used AddToDB earlier, only this time, we want to query out all the people in the database and update the one we choose. First, let's output those names to the screen.

<?php // Instantiate the class $SQL = new MyDatabase; // Connect to the database $SQL->Connect(); ? > <html> <head> <title>Using UpdateDB</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> </head> <body> <table width="500" border="0" cellspacing="0" cellpadding="3"> <tr bgcolor="#CCCCCC"> <td width="121">Name</td> <td width="30">Age</td> <td width="79">Gender</td> <td width="246">Phone</td> </tr> <?php $results = mysql_query("SELECT id,name,age,gender,phone FROM my_tbl WHERE (1) ORDER BY name") or die("MYSQL ERROR: ".mysql_error()); if($row=mysql_fetch_assoc($results)) { do { ? > <tr> <td><a href="<?php echo $_SERVER['PHP_SELF'];

?id=<?php 
   echo $row["id"]; 
"><?php echo 
   $row["name"]; </a></td>
   <td><?php echo $row["age"]; </td>
   <td><?php echo $row["gender"]; </td>
   <td><?php echo $row["phone"]; </td>
   </tr>
  <?php
 }
 while ($row=mysql_fetch_assoc($results));
}
else
{
 ? >
  <tr>
  <td colspan="4">No data</td>
  </tr>
 <?php
}
? >
</table><br>

I trust that the above code needs no in-depth explanation. It's simply connecting to the database using the MyDatabase class and pulling an associative array while looping through the results. Now let's alter our HTML form from above to be used for updating and not inserting.

<?php if($_GET['id']) { $results = mysql_query("SELECT name,age,gender,phone FROM my_tbl WHERE id=".$_GET['id']) or die("MYSQL ERROR: ".mysql_error()); $row=mysql_fetch_assoc($results);


 <form action="<?php echo 
 $_SERVER['PHP_SELF']; " 
 method="post" name="myForm">
  <p>Name: <input name="name" 
  type="text" value="<?php echo
  $row["name"]; "></p>
  <p>Age: <input name="age" type="text"
  value="<?php echo $row["age"]; "></p>
  <p>Gender:<br>
     <input name="gender"
  type="radio" value="MALE"<?php echo 
  ($row["gender"] == 'MALE') ? 
  ' checked' : ''; > Male<br>
     <input type="radio" 
  name="gender" value="FEMALE"<?php echo 
  ($row["gender"] == 'FEMALE') ? 
  ' checked' : ''; > Female</p>
  <p>Phone Number: <input name="phone" 
  type="text" value="<?php echo 
  $row["phone"]; "></p>
  <p><input name="action" type="submit"
  value="UpdateDB">
  <input name="id" type="hidden" 
  value="<?php echo $_GET["id"]; "></p>
</form>
<?php
}
$SQL->Disconnect();
? >
</body>
</html>

The only real difference here is the addition of:

This is just a hidden form field to pass the id. You will also notice we are echoing out the data in the form fields like so:

value="<?php echo $row["age"]; ?>"

Alright, now we have got to make this beast actually update the data in the database table.

Next: You Wouldn't Have to Update It Had You Gotten It Right the First Time >>