Hello, i hope you can help.
I have two tables, listings and property_session
I'm trying to write some code which basically says:
If ListingID is in property session table, then display one image if not then display another.
Can anyone help?
If ListingID is in property session table, then display one
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- AbraCadaver
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Re: If ListingID is in property session table, then display one
Assuming MySQL and that you have the ID in $ListingID:
Code: Select all
$result = mysql_query("SELECT * from property_session WHERE ListingID = '$ListingID'");
if (mysql_num_rows($result) > 0) {
//found
//echo an image
} else {
//not found
//echo another image
}mysql_function(): WARNING: This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQLextension should be used. See also MySQL: choosing an API guide and related FAQ for more information.