[SOLVED] php search form

hakmir · October 14, 2008

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 :lol: ). 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)

<!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 experience ' . '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)

<!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.

[attachment deleted by admin]

MadTechie · October 15, 2008

Heres a VERY basic example of what i think you want..

<?php
if(!isset($_POST['submit'])) die("FAILED: nothing selected");
$dbcnx = @mysql_connect('localhost', 'root', '');
if (!$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 experience ' . 'database at this time.</p>');
}
$typeoffood = $_POST['typeoffood'];
$country = $_POST['country'];
$sql = "SELECT * FROM recipeform WHERE country='$country' AND typeoffood='typeoffood'";
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result))
{
	echo "name".$row["name"]."<br>";
	echo "country".$row["country"]."<br>";
	echo "typeoffood".$row["typeoffood"]."<br>";
	echo "yourrecipe".$row["yourrecipe"]."<br>";
}
?>

hakmir · October 15, 2008

Thanks a lot for the answer. I really appreciate. But when I use the code that you gave me it says:

FAILED: nothing selected

I don't understand. Obviously it doesn't get the values from "searchrecipe.php" page. (id=country and id=typeoffood)

MadTechie · October 15, 2008

opps, change

 if(!isset($_POST['submit'])) die("FAILED: nothing selected");

to

 if(!isset($_POST['Search'])) die("FAILED: nothing selected");

hakmir · October 15, 2008

Awesome. It works perfect. I just had to correct the part that you forgot to put "$" sign to "typeoffood". Thanks a lot. Do you have a paypal account that I can send money. It is not much but just to show my appreciation.

MadTechie · October 15, 2008

I do but a thanx will do

I'm happy to help

hakmir · October 15, 2008

Last question, when I press search, it just shows one result. It doesn't show all the recipies in that criteria (form example, deserts from USA) ??

There are more than one recipe in that category but it just gives me the first one when I press search. It doesn't give me others.?

hakmir · October 15, 2008

That's ok, I will just work on it. Thanks a lot. After I finish this project I will prepare an educational file to teach people (who are very new to php like me) how to make a simple php mysql form from A to Z. Thanks again.

Andy17 · October 15, 2008

Last question, when I press search, it just shows one result. It doesn't show all the recipies in that criteria (form example, deserts from USA) ??

There are more than one recipe in that category but it just gives me the first one when I press search. It doesn't give me others.?

The code for getting all of the results looks good to me. I even tested it just to make sure I didn't miss a simple mistake but it does work for me (with one of my tables - same concept, different content).

PS - You will probably want your results lined up nicely, so I put them into a table that will automatically adjust to the amount of returned rows for you:

<?php
   if(!isset($_POST['submit'])) die("FAILED: nothing selected");
   $dbcnx = @mysql_connect('localhost', 'root', '');
   if (!$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 experience ' . 'database at this time.</p>');
   }
   $typeoffood = $_POST['typeoffood'];
   $country = $_POST['country'];
   $sql = "SELECT * FROM recipeform WHERE country='$country' AND typeoffood='$typeoffood'";
   $result = mysql_query($sql);

if ($result)

{

echo '<table border="0" width="100%"><tr><td width="25%"><b>Name</b></td><td width="25%"><b>Country</b></td><td width="25%"><b>Type</b></td><td width="25%"><b>Recipe</b></td></tr>';

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

        echo '<tr><td width="25%">' . $row['name'] . '</td><td width="25%">' . $row['country'] . '</td><td width="25%">' . $row['type'] . '</td><td width="25%">' . $row['recipe'] . '</td></tr>';

   }

echo '</table>';

}

?>

Should work fine unless I made some silly mistake somewhere (should be easy to fix if I did, anyhow).

hakmir · October 15, 2008

I solved the problem. Here is the code:

<?php


  if(!isset($_POST['Search'])) die("FAILED: nothing selected");
   $dbcnx = @mysql_connect('localhost', 'root', '');
   if (!$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>');
   }
   $typeoffood = $_POST['typeoffood'];
   $country = $_POST['country'];
   $sql = "SELECT * FROM recipeform WHERE country='$country' AND typeoffood='$typeoffood'";
   $result = mysql_query($sql);
   
   
   
  
   	echo "HERE IS ALL THE $typeoffood RECIPES FROM $country"."<br>"."<br>";   
     
      
      
      //my stuff
        
     while ($rec = mysql_fetch_array($result)) {
      $recname = htmlspecialchars($rec['name']);
      $reccountry = htmlspecialchars($rec['country']);
      $rectypeoffood = htmlspecialchars($rec['typeoffood']);
      $recyourrecipe = htmlspecialchars($rec['yourrecipe']);
      
      
      
		 echo "Name: <td>$recname</td>\n"."<br>";
		 echo "Country: <td>$reccountry</td>\n"."<br>";
		 echo "Typeoffood: <td>$rectypeoffood</td>\n"."<br>";
		 echo "Your Recipe: <td>$recyourrecipe</td>\n"."<br>"."<br>";
	  
  	echo( '<a href="searchrecipe.php">Search Recipe</a>' )."<br>";
	echo( '<a href="submitrecipe.php">Submit Recipe</a>' )."<br>"."<br>";
	  
	  
	  }	   

		 //mystuff   

?>

hakmir · October 15, 2008

Belive me I almost gave up learning php. But now I finished my first project with your help. Thanks a lot.

Andy17 · October 15, 2008

Looks to me as if you have some syntax errors in there (in your echos), but if you say it works for ya...

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[SOLVED] php search form

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