[SOLVED] Help!

[Trium918]

Ok, the code below is selecting an image for the database

after the user selects an image name from the drop down

menu.

 

Before the user selects an option echo "No results found";

is displayed. I am trying to a have default value/image to

display at first glance. The page should be set with an image

before the user comes to the page.

 

What am I looking over?

 

 

<?php
  // initialize variable $searchType
  $searchtype = $_POST['searchtype'];
  // connect
  $sql = "SELECT * FROM members_images WHERE images_name='$searchtype'";  													
  if ($result = mysql_query($sql)) {
    if (mysql_num_rows($result)) {
      while ($row = mysql_fetch_assoc($result)) {
    $images_name = $row['images_name'];
        echo "<img src=\"$images_name\" height=\"127\" width=\"170\" alt=\" \" />";
      }
    } 
else {
      echo "No results found";
    }
  } 
  else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }
?>

 

Code for drop down menu

<?php
$query2 = "SELECT DISTINCT images_name FROM members_images";
$result2 = mysql_query($query2);
$num_results2 = mysql_num_rows($result2);

echo "<form name=\"auto\" action=\"edit_photo.php\" method=\"post\">";
echo "<select name=\"searchtype\" onChange=\"auto.submit();\">
<option>Select Images</option>";


for ($i = 0; $i < $num_results2; $i++) {
$row = mysql_fetch_array($result2);
   echo "<option value=\"$row[images_name]\">$row[images_name]</option>\n";
} 
   echo "</select>";
    //echo "<input type=\"submit\" value=\"Submit\" />";
echo "</form>";
?>

[Trium918]

*bump*

[marcus]

<?php
// initialize variable $searchType
$searchtype = $_POST['searchtype'];
// connect
$sql = "SELECT * FROM members_images WHERE images_name='$searchtype'";
$result = mysql_query($sql) or die(mysql_error());
	if (mysql_num_rows($result) > 0) {
		while ($row = mysql_fetch_assoc($result)) {
		$images_name = $row['images_name'];
		echo "<img src=\"$images_name\" height=\"127\" width=\"170\" alt=\" \" />";
		}
	} else {
	echo "No results found";
	}
?>

 

Give that a go.

[dj-kenpo]

or catch it right away at the top

 

<?
$searchtype = $_POST['searchtype'];
if ($searchtype == ""){
//defualt
$sql = "SELECT * FROM members_images LIMIT 1";

}else{
// connect
$sql = "SELECT * FROM members_images WHERE images_name='$searchtype'";
}

?>

[Trium918]

Thanks to both! How can I send the name of the

image to another page? I am having trouble doing

that.