PHP SQL Select From Where - php

I am having some difficulty running some SQL code.
What I am trying to do is, find a row that contains the correct username, and then get a value from that correct row.
This is my SQL in the php:
mysql_query("SELECT * FROM users WHERE joined='$username' GET name")
As you can see, it looks for a username in users and then once found, it must GET a value from the correct row.
How do I do that?

You need some additional PHP code (a call to mysql_fetch_array) to process the result resource returned by MySQL.
$result = mysql_query("SELECT name FROM users WHERE joined='$username'");
$row = mysql_fetch_array($result);
echo $row['name'];

mysql_query("SELECT `name` FROM users WHERE joined='$username' ")
Just select the right column in your 'select clause' like above.
Edit: If you are just starting out though, you might want to follow a tutorial like this one which should take you through a nice step by step (and more importantly up to date functions) that will get you started.

mysql_query("SELECT name FROM users WHERE joined='$username'")

$q = mysql_query("SELECT * FROM users WHERE joined='$username'");
$r = mysql_fetch_array($q);
$name = $r['user_name']; // replace user_name with the column name of your table

mysql_query("SELECT name FROM users WHERE joined='$username' ")
Read documentation : http://dev.mysql.com/doc/refman/5.0/en/select.html

Related

How to store a PHP variable from a SQL table INT camp

This is my table:
All I want to do is to obtain the '75' int value from the 'expquim' column to later addition that number into another (75+25) and do an UPDATE to that camp (now it is 100).
Foremost, there are dozens of ways to accomplish what you want to do. If you're querying the table, iterating over results and doing some conditional checks, the following will work for you. This is pseudo code... Check out the function names and what parameters they require. $db is your mysqli connection string. Obviously replace tablename with the name of your table. The query is designed to only select values that are equal to 75. Modify the query to obtain whatever results you want to update.
This should get you close to where you want to be.
$query = "SELECT * FROM tablename WHERE idus='1'";
$result = mysqli_query($db, $query);
while($row = mysqli_fetch_assoc($result)) {
if($row['expquim'] == 75){
$query2 = "UPDATE tablename SET expquim='".$row['expquim']+25."' WHERE idus='".$row['idus']."' LIMIT 1 ";
$result2 = mysqli_query($db,$query2);
}
}

php/mySQL: querying records by id from array?

i want to query several records by id like:
$ids = array(10,12,14,16,....);
the query would be something like:
select * from users where id=10 or id=12 or id=14 or id=16 ..
is it possible to query it in a more comfortable way like (compared to php):
select * from user where in_array(id, array(10,12,14,16))
thanks
You can use IN instead of OR clauses
select * from users where id IN (put_your_array_here)
Example:
select * from users where id IN (10,12,14,16);
Note:
According to the manual for MySQL if the values are constant IN
sorts the list and then uses a binary search. I would imagine that
OR evaluates them one by one in no particular order. So IN is
faster in some circumstances.
Related post
Try this.
$id = array(10,12,14,16,....);
$ids = join("','",$id);
$sql = "select * from user where id IN ('$ids')";
OR
$ids = implode(',', $id);
$sql = "select * from user where id IN ($ids)";
You can do it like that using implode in PHP:
"select * from users where id in '".implode("','", $ids)."'"
Please be sure that your ids are safe though

PHP/MSSQL.. Select ID based on username, insert to table based on ID?

I have looked all over and at tons of code and examples.. This is such a small bit of code but I just can't seem to get it to work.
I have dbo.accounts which contains the id, username, password, createtime..
I have a simple form, you type in the username, and I need the select query to return the ID based on the username.
$result = mssql_query('SELECT id FROM dbo.account WHERE name = $username');
The dbo.gamemoney table will just insert some hardcoded info such as an amount of coins for the game..
My problem is that if I use a query as ID = 123, it works, but when I try to grab the id of dbo.accounts by using the username, I get nothing back.
I know it has to be something small, But I have tried to figure it out for so many hours now that I'm honestly lost..
Thanks for your time,
Chris
Since, $username is string type, you have to enclose it in quotes.
$result = mssql_query("SELECT id FROM dbo.account WHERE name = '$username'");
As a better practice would suggest use a try-catch scenario so that you get the exact error log. Try -
$result = mssql_query('SELECT id FROM dbo.account WHERE name = "'.$username.'"') or die('MSSQL error: ' . mssql_get_last_message());
Thanks everyone for the help!
I was able to get it working. Now I'll make sure it's the right way. I had forgot to add,
while($row = mssql_fetch_array($result)) {
$id = $row['id'];
$ip = $row['ip'];
}
Thats why the id was blank. I was missing some code.
Chris

Can't access row with 'fieldName' using MAX() in PHP MYSQL

I have small PHP script which has
$query = "SELECT MAX(id) FROM `dbs`";
//query run
$row = mysql_fetch_array($result);
$val = $row[0];
Which runs fine, but I want to understand why i can't access the row with the fieldname, like if i have this
$query = "SELECT id FROM `dbs`";
i am able to use the folowing
$val = $row['id'];
but whenever i use this MAX() function, i have to change to
$val = $row[0];
to access the values
I have no clue about this. Any help would be appreciated. Thankss
You need to give it an alias:
<?php
$query = "SELECT MAX(id) AS `id` FROM `dbs`";
//query run
$row = mysql_fetch_array($result);
$val = $row['id'];
Edit:
To explain this it's probably best to show an example of a different query:
SELECT MAX(`id`) AS `maxId`, `id` FROM `dbs`
Using the above it will return as many rows are in the table, with 2 columns - id and maxId (although maxId will be the same in each row due to the nature of the function).
Without giving it an alias MYSQL doesn't know what to call it, so it won't have an associative name given to it when you return the results.
Hope that helps to explain it.
SELECT MAX(id) AS myFieldNameForMaxValue
FROM `dbs`
and then
$row = mysql_fetch_array($result);
$val = $row['myFieldNameForMaxValue'];
If you run this query on mysql commandline you'll see that the field name returned by mysql is MAX(id). Try running on phpmyadmin and you'll see the same. So if you try $row['MAX(id)'] it'll work. When using a mysql function, it gets added to the name, so use an alias, like other said here, and you're good to go: SELECT MAX(id) AS id FROM dbs. Also, never forget to use the ` chars, just in case you have some columns/tables with reserved names, likefrom`.

MySQL Query Isnt Returning A Result

i got a fairly simple layout going and for the life of me i cant figure out why this returns nothing:
<?php
// Gets A List Of Comic Arcs
$result = mysql_query("SELECT * FROM ".$db_tbl_comics." GROUP BY ".$db_fld_comics_arc." ORDER BY ".$db_fld_comics_date." DESC LIMIT 20");
while ($comic = mysql_fetch_array($result)) {
// Now Go Back And Count Issues For Each Comic Arc Above
$result22 = mysql_query("SELECT * FROM ".$db_tbl_comics." WHERE ".$db_fld_comics_arc."=".$comic[$db_fld_comics_arc]);
$total_issues = mysql_num_rows($result22);
echo $total_issues;
}
?>
No other query is refered to as $result22.
$comic[] has already been defined in the previous query.
echo mysql_error($result22); returns no errors.
Let me know if you need any other info.
I am assuming that the column $db_fld_comics_arc is a string.
Change:
$result22 = mysql_query("SELECT * FROM ".$db_tbl_comics." WHERE ".$db_fld_comics_arc."=".$comic[$db_fld_comics_arc]);
To:
$result22 = mysql_query("SELECT * FROM ".$db_tbl_comics." WHERE ".$db_fld_comics_arc."='".$comic[$db_fld_comics_arc]."'");
Am I wrong? If so, let me know the table structure, and what your error reporting is set to.
Also, could you let us know the purpose of your SQL? It may also be possible to put the data together in one query, instead of looping sql queries through, and using data from a first query.
Maybe it is because $db_fld_comics_arc is in $comic[$db_fld_comics_arc]
if both are the same then you should try replacing $db_fld_camics_arc with $comic[$db_fld_comics_arc].

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