MySQL noob question

bigger

How can I get information out of a table and display ever result? I get resource id #2 or a blank result when I was array? Any code example would be most helpful 🆒 Thanks!

[deleted]

http://www.yapf.net/faq.php?cmd=viewitem&itemid=12

bigger

Alright, I tried what they said there too, again I received a blank return. What's up with this?

I thought it'd help if I put one of the many different kinds of queries I've tried, most of them have all returned just a blank result:



$query = "SELECT `field1` FROM `theTable` LIMIT 0, 30";


$result = mysql_query(query);


echo ("$result");

This gave me the same thing as what the link you gave me did, a blank result.

[deleted]

It seems there were a few typos in the content, but that's fixed now.

Read the article, because it explains how to use mysql with PHP. It's not complex but it's one of those things you need to know to get it to work.

bigger

I know how to do PHP and MySQL, the problem is not my lack of knowledge, it just seems no matter what function I use to try and get the data, it just comes back blank, I've tried fetch_array, fetch_object, fetch_row, but all of them returned blank results. Could this be a block in my host's table?

[deleted]

" I know how to do PHP and MySQL, the problem is not my lack of knowledge,"

If the code you postyed here represents your level of knowledge then I'd say you have a long way to go.

Post your real code and we'll see what the problem is.

Obviously it not a 'block' on the table because if it was, you'd get an 'access denied' message.

bigger

I've been doing more now, I took your example, copied it, pasted it, put in the nesscary changes, and it even brought back a blank result, it's in a standalone file, and all it's suppose to do is put what it gets, and it puts nothing, just blank, empty space. What's going on?

TalonFinsky

Just an example:

<?php
$db = mysql_connect(servername.com, user, pass)

mysql_select_db("dbname");

$query = "SELECT * FROM table WHERE field = ".$compare
$result = mysql_query($query, $db);
$num = mysql_num_rows($result);

for ($i = 0; $i < $num; $i++)
{
  $row = mysql_fetch_array($result);
  $content = $row[stuff]
  /****
     display how you want
  ****/
}

?>

A little vague, but should get you pointed somewhat in the right direction.

bigger

That too, brings a blank result. What the heck could this be?

TalonFinsky

Paste in the exact code you used on that last try. (just the query snippet is fine)

bigger

<?php
$db = mysql_connect(localhost, username, password);

mysql_select_db("thedatabase");

$query = "SELECT 1 FROM picposts WHERE field = ".$compare;
$result = mysql_query($query, $db);
$num = mysql_num_rows($result);

for ($i = 0; $i < $num; $i++)
{
$row = mysql_fetch_array($result);
$content = $row[stuff];

echo $content;
}

[deleted]

You select only '1' from the database, yet you try yto print a field called 'stuff'.

you'll have better results if you name the field you are selecting, or give the selected values a name:

SELECT 1 AS stuff FROM 'picposts

Also, as I stated in the article, it is very important to do proper error-checking and reporting. Without it, you will spend days fixing 'silly' bugs like these.

bigger

1 is the name of the field I want to print. And there is no field called stuff. I get no errors, just a blank result, neither does your example script work, it just gives a blank result too. And I put in the AS stuff, and still doesn't bring anything back, just a blank result. Now, the odd thing is, using your method on the example page, I am calling something out of another table, but then when I copied over exactly what I had going there, and just changed the query, it didn't work either. So, I'm clueless.

ultraslacker

Make your life easier by learning how to debug - I'm not even talking about error reporting, just simple echo's that let you know where the problem might be.

Your echo statement is only helpful if you get a result. Add a label - echo "Content: $content"

Fix your query - 1 is not a valid column name.

Echo the query to the screen and try it in a mysql client.

Echo numrows - I suspect you are getting 0 results.

bigger

I tried my query in phpmyadmin, and it brought up the result I wanted. But good idea with the numrow, I didn't even think of doing that.

--- Later ---

Alright, I echoed the number of rows I was returning, and that said 1. So, I am getting a result, it's just not displaying I guess then?

--- Later Again ---

I did the other things too, I changed the field name, still didn't bring anything, and I echoed the query, and tried it again in phpmyadmin just to be sure, and it returned what I wanted, the text in the field.