Pulling info out of a table and into a form...

mysql_query

Hello.

I have a regular HTML form, in which I want to create a drop-down box. In the drop down box, I want it to have ids of rows I have in a mysql table. When the form is submitted, I want the results posted in ANOTHER table with that id as one of the columns. From what I understand, this is mostly HTML (the form is made in html, not php, the submitter is a diffrent file.)(it's 'ation=submit.php", not "ation=php_self")
Any help (coughscode 😉 ) is appreciated.

Thanks.

psy

mysql_connect(you know what to do);
mysql_select_db(db);
$query="SELECT * FROM table";
$r=mysql_query($query);

if($r && mysql_num_rows($r)>0)
{
 echo "<select name='id'>\n";
 for($i=0;$i<mysql_num_rows($r);$i++)
 { 
   $arr=mysql_fetch_array($r);
   echo "<option value=" . $arr['id'] . ">$arr['id']";
 }
 echo "</select>";
}else
 echo "ut-oh";

that puts all the IDs in the table into a dropdown box.. see next post for the submit thing

psy

ok, after posting to whatever, search out the ID and grab it like this

<?
if(isset($_POST['id']))
{
 mysql_connect(again);
 mysql_select_db(db);
 $query="SELECT * FROM table WHERE id=" . $_POST['id'] . " LIMIT 1";
 $r=mysql_query($query);
 if($r){
  $arr=mysql_fetch_array($r);
?>
  <!-- FORM(at) however you want it to be viewed -->
  <!-- do this with each $arr['key'] to get what you want -->
  <input type=text name="username" value="<?=$arr['name']?>">

<?
}// end $r if
}// end _post if
?>

thats the basic idea (at least for the problem i think you gave me 🙂 i could have read it completely wrong..)

mysql_query

Hi. Thanks for taking your time to answer.

I don't really get with this does, can you please post comments in the script as to what functions does what.

Thanks.

psy

the fist script will put all the 'id's from the 'table' into a select box.. put that script somewhere between <form action='submit.php' method='post'> tags, and when you submit it, it will pass the 'id' var

then the 2nd post (submit.php) checks for $_POST['id'], if found, it will then bringup that row from the databse and print out (however you want to format it) the values

mysql_query

Ok, thanks a lot, I'll try it.

kilbey1

And how would you edit that code to show what's currently selected for that ID in the database?

-Eve
eatley@tribweb.com

sanu

$query="SELECT * FROM table WHERE id=" . $_POST['id'] . " ;
$r=mysql_query($query);

while($arr=mysql_fetch_row($r)){
print("$arr[1]<br>$arr[2]<br> ");
}

Dynamic2000

Hi,

I'm trying to use this script as well, but I always get the following error message:

Parse error: parse error, expecting T_STRING' orT_VARIABLE' or `T_NUM_STRING' in C:\apache\htdocs\dropdowntest.php on line 34

where line 34 is the following line: echo "<option value=" . $arr['id'] . "/>$arr['id']";

This is the form part of the script:

<?php
mysql_connect("localhost","user","passw");
mysql_select_db("testdb");

$query="SELECT * FROM testtable";
$r=mysql_query($query);

if($r && mysql_num_rows($r)>0)
{
echo "<select name='id'>\n";
for($i=0;$i<mysql_num_rows($r);$i++)
{
$arr=mysql_fetch_array($r);
echo "<option value=" . $arr['id'] . "/>$arr['id']";
}
echo "</select>";
}else
echo "ut-oh";
?>

</select></form>

I hope somebody can help me! Thanks

Dynamic2000 😕

Dynamic2000

I already solved the problem guys! So no need to reply....

😃

the_link

hi hhow did u solve ur problem plz coz i got the same error message!

thanx

masterKAHN

Hi mysql_query!

You posted a post to your thread where you wrote that you couldn't understand what psy wrote (in his code). Well take your time reading the code understanding what is what. Programming is difficult and only the best can just read the code like a novell. Therefore like I and thousands of other beginners, intermediate and even some very good programmers read the code carefully even more times to get it right. Thats the best exercising you can do.

Dynamic2000

Sorry for the late reaction. What I did is replace the this line:

echo "<option value=" . $arr['id'] . "/>$arr['id']";

with

echo "<option value='".$arr['id']."'>".$arr['id']. "</option>"

Then it worked!

Good luck!

hoham

Thanks to all that respond to everyones questions.

I search the net for all answers to my questions. I find solutions and hints from everyone that ever asks a question, or answers.

While Books & documentation are great resources, its nice to see how actual problems are solved.

I'm always looking for a hint, so I can try and solve the problem myself. Sometimes its just one of those damn missing ( ' ) or similar missed punctuation, or I'm completely off trying to write newbie code in pages, that in reality, only takes a few lines.

I thank every one. Those who ask, and those that take the time to share their answers -

dan