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Dynamic form population. (Almost Solved)


tet3828

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In the following script...Why did the coder Leave PHP ?> Make the <select> statement and go back into it <?php?

would this code still work if it stayed in PHP and used the echo "<select name="blah">.....???

<?php
php require("config.php");
$result = mysql_query("SELECT * FROM spell");
?>

<select name="wordlist">
<option value="1" selected>Select Word</option>

<?php
   while($row = mysql_fetch_array($result)){
      echo "<option value=\"$row['words']\">$row['words']</option>\n";
   }
?>
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It doesnt matter, you can do that.
This is the same thing as-

[code]<?php
php require("config.php");
$result = mysql_query("SELECT * FROM spell");

echo '<select name="wordlist">\n<option value="1" selected>Select Word</option>';

   while($row = mysql_fetch_array($result)){
      echo "<option value=\"$row['words']\">$row['words']</option>\n";
   }
?>[/code]

Orio.
what would cause these lines to have a Parse error:

parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/content/t/e/t/tetunity/html/shell/data/edit.php on line 202???

    while($row = mysql_fetch_array($result)){
echo "<option value=\"$row['itemCat']\">$row['itemCat']</option>\n";
    }
Maybe my question was misunderstood or my script is inncorrect.

for my 'products' database is setup as follows:

each "product" has "itemName, itemDesc, itemCat.....
I was hoping to have a dropdown menu filled with ANY values that maybe listed in the itemCat column.
for example even if I went to myphpadmin and added a new item with a new catagory that catagory
would appear in my dropdown. for some reason this script isn't populating my dropdown with anything
other then the "select a catagory" option. any suggestions as to why this might be happening????

echo "<select name='newName'>";
      echo "<option value='1' selected>Select a Catagory</option>";
      echo "<input type='submit' name='submitfour' value='Submit' /><br />";
        $result = mysql_query("SELECT itemCat FROM products");
      while($row = mysql_fetch_array($result)){
echo "<option value=\"".$row['itemCat']."\">".$row['itemCat']."</option>\n";
          }

Try this:

[code]<?php
 
echo "<option value=\"".$row['itemCat']."\">".$row['itemCat']."</option>\n";
$result = mysql_query("SELECT DISTINCT itemCat FROM `products`"); 
while($row = mysql_fetch_array($result))
{
echo "<option value=\"".$row['itemCat']."\">".$row['itemCat']."</option>\n";
}

echo "</select>";

?>[/code]

Orio.
Almost there! the Dynamic form is [i][b]almost[/b][/i] ERROR free.
Thanks once again [b]Orio you're a Genius[/b]! Thanks once again for all the time you devote to this fourm

The only problem I am having now is: the loop is displaying one extra line in my dripdown menu. The extra line is blank. What is causing this? and how can I eradicate it?


echo "<select name='newName'>";
echo "<option value='1' selected>Select a Catagory</option>";
echo "<option value=\"cat".$row['itemCat']."\">".$row['itemCat']."</option>\n";
$result = mysql_query("SELECT itemCat FROM `products`"); 

while($row = mysql_fetch_array($result)) {
    echo "<option value=\"".$row['itemCat']."\">".$row['itemCat']."</option>\n";
}
echo "</select>";
echo " <input type='submit' name='submitfour' value='Submit' /><br /><br />";
[code][codegecho "<select name='newName'>";
echo "<option value='1' selected>Select a Catagory</option>";
echo "<option value=\"cat".$row['itemCat']."\">".$row['itemCat']."</option>\n"; //<= this is extra line remove it
$result = mysql_query("SELECT itemCat FROM `products`"); 

while($row = mysql_fetch_array($result)) {
    echo "<option value=\"".$row['itemCat']."\">".$row['itemCat']."</option>\n";
}
echo "</select>";
echo "  <input type='submit' name='submitfour' value='Submit' />

";[/code]

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