php mysql search form
Posted: Tue Oct 14, 2008 1:26 pm
Hi Guys,
I made 3 files:
submitrecipe.php
searchrecipe.php
reciperesult.php
I don't have any problem submitting imformation with "submitrecipe" to mysql database. I managed to make "searchrecipe.php" too (I think so
). but I don't know how to make "reciperesult.php to show the results. I just prepared an image on html form to show you what I am trying to do (reciperesults.jpg). I attached other files too which I don't have problems with them (submitrecipe.jpg and searchrecipe)
Here is "submitrecipe.php" (I don't have problem with this form)
Here is "searchrecipe.php" (I don't have problem with this one either. As far as I know)
Now I don't know how to make this file "reciperesult.php"
Can anybody help. Thanks a lot.
I made 3 files:
submitrecipe.php
searchrecipe.php
reciperesult.php
I don't have any problem submitting imformation with "submitrecipe" to mysql database. I managed to make "searchrecipe.php" too (I think so
Here is "submitrecipe.php" (I don't have problem with this form)
Code: Select all
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form id="form1" name="form1" method="post" action="submitrecipe.php">
<label><strong>SUBMIT YOUR RECIPE</strong><br />
<br />
Name<br />
<input type="text" name="yourname" id="yourname" />
</label>
<p>Country<br />
<label>
<select name="yourcountry" id="yourcountry">
<option>USA</option>
<option>CANADA</option>
<option>ENGLAND</option>
</select>
</label>
</p>
<p>Type of Food <br />
<label>
<select name="typeoffood" id="typeoffood">
<option>DESERT</option>
<option>ENTREE</option>
<option>SPICY</option>
</select>
</label>
</p>
<p>Your recipe<br />
<label>
<textarea name="yourrecipe" id="yourrecipe" cols="45" rows="5"></textarea>
</label>
</p>
<p>
<label>
<input type="submit" name="Submit" id="Submit" value="Submit" />
</label></p>
</form>
<?php
$dbcnx = @mysql_connect('localhost', 'root', '');
if (!$dbcnx) {
exit('<p>Unable to connect to the ' . 'database server at this time.</p>');
}
if (!@mysql_select_db('recipetime')) {
exit('<p>Unable to locate the recipe ' . 'database at this time.</p>');
}
if (isset($_POST['yourrecipe'])):
// A new recipe has been entered
// using the form.
$yourname = $_POST['yourname'];
$yourcountry = $_POST['yourcountry'];
$typeoffood = $_POST['typeoffood'];
$yourrecipe = $_POST['yourrecipe'];
$sql = "INSERT INTO recipeform SET
name='$yourname',
country='$yourcountry',
typeoffood = '$typeoffood',
yourrecipe='$yourrecipe'";
if (@mysql_query($sql)) {
echo '<p>New recipe added</p>';
} else {
exit('<p>Error adding new recipe: ' . mysql_error() . '</p>');
}
?>
<?php endif; ?>
</body>
</html>
Here is "searchrecipe.php" (I don't have problem with this one either. As far as I know)
Code: Select all
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<?php $dbcnx = @mysql_connect('localhost', 'root', '');
if (!$dbcnx) {
exit('<p>Unable to connect to the ' .
'database server at this time.</p>');
}
if (!@mysql_select_db('recipetime')) {
exit('<p>Unable to locate the recipe ' .
'database at this time.</p>');
}
?>
<form id="form1" name="form1" method="post" action="reciperesults.php">
<label><strong>SEARCH RECIPE</strong><br />
<br />
</label>
<p>Country<br />
<label>
<select name="country" id="country">
<option>USA</option>
<option>CANADA</option>
<option>ENGLAND</option>
</select>
</label>
</p>
<p>Type of Food <br />
<label>
<select name="typeoffood" id="typeoffood">
<option>DESERT</option>
<option>ENTREE</option>
<option>SPICY</option>
</select>
</label>
</p>
<p>
<label>
<input type="submit" name="Search" id="Search" value="Search" />
</label>
</p>
</form>
</body>
</html>
Now I don't know how to make this file "reciperesult.php"
Can anybody help. Thanks a lot.