inserting an id from one table to another

[average american]

I need help this may be easy i don't know...

 

 

I need to link my two table together.  The first table(listings) has a row ID (primary and auto inc) my second table (images) has a row LISTING_ID.  I cannot get the last id from the listings table to get inserted on to the images table.

 

I have the part of my code below:

 

mysql_query("LOCK TABLES listings, images WRITE");

 

mysql_query("INSERT INTO images (name, size, type, content ) ".

"VALUES ('$fileName', '$fileSize', '$fileType', '$content')");

$result = mysql_query("SELECT MAX(ID) AS LAST_ID FROM listings");

 

$id = mysql_fetch_array($result);

mysql_query("UNLOCK TABLES");

 

echo "Last ID#" . $id[LAST_ID];

echo "<br>File $fileName uploaded<br>";

 

 

the images load and i am able to see the ID via the echo "Last ID#" .$id[LAST_ID'] part of the code.  But, nomater what I have tried i cannot get that id to be posted in the listing_id  of the images table.

[kireol]

where are you inserting into your listings table?

[average american]

from another page....

i would like to have this all on the same page but it was way to confusing.

 

this is the code for the listings

 

<?php

if (isset($_REQUEST['Submit']))

{

 

# THIS CODE TELL MYSQL TO INSERT THE DATA FROM THE FORM INTO YOUR MYSQL TABLE

$address=mysql_real_escape_string($_POST['address']);

$city=mysql_real_escape_string($_POST['city']);

$state=mysql_real_escape_string($_POST['state']);

$bed=mysql_real_escape_string($_POST['bed']);

$bath=mysql_real_escape_string($_POST['bath']);

$garage=mysql_real_escape_string($_POST['garage']);

$rent=mysql_real_escape_string($_POST['rent']);

$description=mysql_real_escape_string($_POST['description']);

$zip=mysql_real_escape_string($_POST['zip']);

$sql = "INSERT INTO listings(address, city, state, bed, bath, garage, rent, description, zip) values ('$address', '$city', '$state', '$bed', '$bath', '$garage', '$rent', '$description', '$zip')";

if($result = mysql_query($sql , $db))

{

echo "Thank you, Your information has been entered into our database";

echo "<a href= '$config_basedir./database entry/addingimages.php'>add images</a>";

}

else

{

echo "ERROR: ".mysql_error();

}

 

}

else

{

?>

<form method="post" action="">

  <p>Address:

    <input name="address" type="text" size="50" maxlength="100">

  </p>

  <p><br>

   

    City:

    <input name="city" type="text" size="30" maxlength="30">

   

    State:

  <input name="state" type="text" size="2" maxlength="2">

    Zip:

  <input name="zip" type="text" size="6" maxlength="5">

  </p>

  <p><br>

   

    Bed:

    <input name="bed" type="text" size="4" maxlength="1">

   

    Bath:

    <input name="bath" type="text" size="4" maxlength="3">

   

    Garage:

    <input name="garage" type="text" size="4" maxlength="1">

   

    Rent: $

    <input name="rent" type="text" size="6" maxlength="4">

  </p>

  <p>    <br>

    Description: <br>

    <textarea name="description" cols="100" rows="5"></textarea>

   

    <br>

    <br>

    <input type="submit" name="Submit" value="Submit">

  </p>

</form>

<p>

 

[kireol]

well, you arent explicitly insertng the ID field.  do you have phpmyadmin or something to verify that that field has data like you expect?

[average american]

i do have phpmyadmin and that is true it is not get inserted.  (that is the listing_id)

 

the listings table is auto id so it is automaticly inserted into the table.

 

i just can not figure out how to get that auto inserted id into the images table and into the listing_id.

 

[redarrow]

you need to use $_GET in the url off the link page.php?id=$id or use a session