Setting up a simple search with PHP5

  RicScott 11:50 17 Sep 2006

I'm trying to set up a simple search that will show only one result..

Basically, someone enters a number and then clicks search, the result should show on another page, picking out the priKey from a mySql database.

I've set up the form, just need help retreiving the data.

  RicScott 10:54 18 Sep 2006

The html form:
<form name="search" method="post" action="results.php">
Enter an ID Number: <input type="text" name="find" />

<input type="hidden" name="searching" value="yes" />
<input type="submit" name="search" value="Search" />

The PHP code on results.php:

//This is only displayed if they have submitted the form
if ($searching =="yes")
echo "<h2>Results</h2><p>";

//If they did not enter a search term we give them an error
if ($find == "")
echo "<p>You forgot to enter an ID Number, Please try again.";

// Connect to DB

// Make the query.
$sql = "SELECT * FROM tbl WHERE ID($field) ='%$find%'";

$result = @mysql_query ($sql); // Run the query.

echo '<table align="center" cellspacing="0" cellpadding="5" border="0" width="100%">

// Fetch and print all the records.
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

echo '<tr bgcolor="' . $bg . '" class="style1">

<td align="left" valign="middle"><img src="' . $row['image_link'] .'" width="401px" height="301px"></td>

<td align="left" align="left" valign="top"><table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr><td align="left" valign="top" class="style5">Property ID: <b> ' . $row['propertyID'] . '</b><br> Priced From: <b>€ ' . $row['price'] . '</b><br>
Type: <b> ' . $row['type'] . '</b><br>Name: <b>' . $row['name'] . '</b><p>
Area: <b> ' . $row['area'] . '</b><br>
Bedrooms: <b> ' . $row['bedrooms'] . '</b> :: Bathrooms: <b> ' . $row['bathrooms'] . '</b><p>
Pool:<b> ' . $row['pool'] . '</b> :: Garden:<b> ' . $row['garden'] . '</b> :: parking:<b> ' . $row['parking'] . '</b><p><br></td></tr>
<tr><td align="left" valign="top" class="style1">' . $row['description'] . '

echo '</table>';

mysql_free_result ($result); // Free up the resources.
mysql_close(); // Close the database connection.

But nothing shows...

  harristweed 11:26 18 Sep 2006

To start with:
$sql = "SELECT * FROM tbl WHERE ID($field) ='%$find%'";

Where is $field set?

why not
$sql = "SELECT * FROM tbl WHERE ID ='%$find%'"; ?


Remove the @ from
$result = @mysql_query ($sql); // Run the query.
$result = mysql_query ($sql); // Run the query.

then you will see the error message, that might help!

  RicScott 12:04 18 Sep 2006

but unfortunately, I still get a blank page..No errors and no programmed errors are returned.

  harristweed 12:54 18 Sep 2006


echo $sql;

and see what the search string looks like

  RicScott 13:30 18 Sep 2006

Still a blank page..this language will be the death of me...

  harristweed 13:48 18 Sep 2006

Could be....
if ($searching =="yes")
echo "<h2>Results</h2><p>";

where's $searching=$_POST[searching]; ?

Therefore $searching doesn't ==yes
do nothing!
so nothing to do, code works!

  RicScott 16:18 18 Sep 2006

Removed the % % from the search which showed a return of -

You forgot to enter an ID number........

Just not showing the result...

I'm getting there.

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