coopster
msg:4316909 | 2:58 pm on May 24, 2011 (gmt 0) |
First, I am going to assume that the table 'domain' mentioned is actually 'alldomain'. If not, you are going to have to rework the information below to add an additional JOIN to the query.
OK, let's say that you have already scrubbed the user-supplied POST data and then pushed it through mysql_real_escape_string() to prepare it for use in your query and you have stored it in a variable called $linkurl. Next is to create your SQL statement ...
$sql =
"SELECT
alldomain.manual_approve,
register.username
FROM alldomain
INNER JOIN register ON(alldomain.userid = register.id)
WHERE
alldomain.domain_name = '$linkurl'"
;
The INNER JOIN is the part that seems to avoid you. See if that works, test it and tweak it if necessary.
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gibsongk55
msg:4317209 | 12:27 am on May 25, 2011 (gmt 0) |
Hi,
Thanks so much for the reply. After the query how do i get the two values moved into a variable say this:
$x_manual_approve = domain.manual_approve
$x_username = register.username
Thanks again,
Gibs
|
jspeed
msg:4317499 | 4:40 pm on May 25, 2011 (gmt 0) |
After coopsters code, you would run the query, and store the results in variables:
$result=mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
$x_manual_approve=$row["alldomain.manual_approve"];
$x_username=$row["register.username"];
}
Then use them as you need:
echo $x_manual_approve;
etc
|
gibsongk55
msg:4317788 | 7:21 am on May 26, 2011 (gmt 0) |
HI JS,
Thanks so much for the reply. I will try that. Sure is confusing reminds me of the old days with arrays and basic programming.
Thanks again,
Gibs
|
gibsongk55
msg:4317811 | 8:53 am on May 26, 2011 (gmt 0) |
Hi,
Thanks for the help guys. I still have a problem. The two fields I am trying to get from the database are coming up empty.
This is the code I have from you:
$sql =
"SELECT
alldomain.manual_approve,
register.UserName
FROM alldomain
INNER JOIN register ON(alldomain.userid = register.Id)
WHERE
alldomain.domain_name = '$link_url'"
;
$result=mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
$x_manual_approve=$row["alldomain.manual_approve"];
$x_username=$row["register.username"];
}
I am not understanding the beginning
SELECT alldomain.manual_approve, register.UserName
the above are two fields in two tables. Then FROM alldomain is only one table. Should that be FROM * ?
Thanks for help.
Gibs
|
coopster
msg:4317878 | 12:06 pm on May 26, 2011 (gmt 0) |
Read my first post again. The part that is escaping you is the JOIN. In order to retrieve values for columns from two different tables you need to join the tables together on their relative key, which in this case you mentioned was the userid. Column names are case-sensitive (looking at register.Username). You wrote it out twice now in two different cases. Check your error logs or dump the query upon error to see if you have issues. Something like this may be helpful during testing ...
$result = mysql_query($sql) or die(mysql_error());
|
gibsongk55
msg:4317893 | 12:56 pm on May 26, 2011 (gmt 0) |
Hi,
Yes i checked the database and that is why i changed the case to match it the field exactly. It is now correct and still the fields come up empty.
I pasted the code into the query in phpmyadmin and this is what I got:
SQL query: Documentation
$sql = "SELECT alldomain.manual_approve, register.UserName FROM alldomain INNER JOIN register ON(alldomain.userid = register.Id) WHERE alldomain.domain_name = '$link_url'";
MySQL said: Documentation
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$sql =
"SELECT
alldomain.manual_approve,
regi' at line 1
Thanks
|
coopster
msg:4318103 | 5:32 pm on May 26, 2011 (gmt 0) |
You cannot paste PHP programming into phpMyAdmin. The query you want to paste to test is the "SELECT ..." part of the code shown here. It is wrapped in quotation marks to make it a string value. The string value is being assigned to the variable $sql. You can use that in your mysql_query() function as demonstrated by jspeed.
Also, $link_url is a variable too. You cannot just paste that into phpMyAdmin. You need to populate that with a value as I mentioned earlier.
|
gibsongk55
msg:4318624 | 3:22 pm on May 27, 2011 (gmt 0) |
Okay I still don't get it.
Okay as you said I copied your code:
SELECT
alldomain.manual_approve,
register.username
FROM alldomain
INNER JOIN register ON(alldomain.userid = register.id)
WHERE
alldomain.domain_name = 'www.example.com'
It returned one record:
manual_approve = 0
username = test@example.com
So that returns the correct value. But the code comes up empty at the end?
Thanks,
Gibs
|
coopster
msg:4319025 | 5:47 pm on May 28, 2011 (gmt 0) |
| But the code comes up empty at the end? |
|
Meaning? The correct result set was returned so are you not printing it out correctly or something else? You'll have to offer more details.
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