I made a portal where i have to fetch result with primary key as id i am using it everywhere storing it in a variable however there is a certain problem that i am facing
$sql="SELECT users from hostelusers WHERE users Like'%$roll%'";
$result = mysql_query($sql) or die(mysql_error());
$rows=mysql_fetch_row($result);
$uid=$rows[0];
$_SESSION['uid']=$uid;
In my code the variable $uid is echoing out values of column 1 ie users however I have selected column 0 which is id(in the table in mysql)
can't think of any mistake there
your mysql query string is wrong . You are using :
$sql="SELECT users from hostelusers WHERE users Like'%$roll%'";
And how do you expect to get id field in the result ??
Moreover you will be having only a single column [field] in your rows[] and that is rows[0]
if you want to use rows[0] to get your id field
Change your query to this
$sql="SELECT id from hostelusers WHERE users Like'%$roll%'";
or this:
$sql="SELECT * from hostelusers WHERE users Like'%$roll%'";
and it will work for sure
Related
When we update a MySQL record with php, we can check if it has effect using:
$mysqli->affected_rows;
But how do I check which column has been modified?
Example, in my table have the columns: id / name / age
In a record we have the data: 1 / Woton / 18
If I send an: UPDATE mytable SET name = 'Woton', age = '20' WHERE id = '1'
Only the age field has changed. How can I determine this?
You cannot directly get the updated columns from the query result.
It can be get from some php query. Firstly we will have to select the row from database which we are going to update in a array variable. Than run the update query for the same row.
Lastly get the same row from database from select query in the new array variable.
Finally we get two arrays.
We can get the updated column with the array_diff_assoc php function.
See the below code for the same.
$sql = "SELECT * from mytable where id=1 limit 1";
$prev = mysqli_fetch_assoc(mysqli_query($conn, $sql));
//Get the column data in the array. Before update.
$sql = "UPDATE mytable SET name = 'Woton', age = '20' WHERE id = '1'";
$conn->query($sql);
// Update data
$sql = "SELECT * from mytable where id=1 limit 1";
$updated = mysqli_fetch_assoc(mysqli_query($conn, $sql));
// Again run the select command to get updated data.
$UpdatedColumns=array_diff_assoc($updated,$prev);
In a different note: If QueryLog has been enabled in the DB then you (or your script in PHP or Python or any) can easily see/read which part of the content has been updated and even you can monitor the DB.
The good news is, even you can target which table, which query etc.
I would save the id of each user in a session for get later name ... in my login system. I would do that on each login.
But the problem: If i Select more columns with the sql i cant fetch a single column.
How i can fetch one column even tough i selected more columns in the sql.
I tried something like that:
$sth->fetch(PDO::FETCH_ASSOC)['password']
But it isn´t working on my code:
$sth = $X['dbh']->prepare("SELECT `password` `id` FROM `users` WHERE `uid` = :uid; ");
if (! $sth->execute(array(
':uid' => $X['param']['uid'],
))) {
//check for right name, passwort ...
}
How i can fetch one column even tough i selected more columns in the sql.
Nohow.
Either select only one column, or fetch all that you have selected. Otherwise it would make no sense to select.
And after fetching an array, you can access its members all right.
$user = $sth->fetch(PDO::FETCH_ASSOC);
$pass = $user['password'];
then later in the code you can use $user['id'] as well.
Hello I have 2 textboxes and i want to give to the user the option to choose one in order to find results. The user can search through the id or the name. My problem is because i use LIKE%field% when the user chooses to search through the id the name field stays empty and returns all the table rows. I want to have results only if the user enters some value in the textbox. This is my sql query. I'm using mysql
"SELECT * FROM properties WHERE ID='$id' OR Name LIKE '%$name%'"
Thank you all
If the user has to select which field to search, you can do:
if ($_POST['search'] == 'id') {
$sql = "SELECT * FROM properties WHERE ID='$id'"
} else {
$sql = "SELECT * FROM properties WHERE Name LIKE '%$name%'"
}
You can do this in a single query (values are checked from the query itself):
"SELECT * FROM properties WHERE ('$id'='' OR ID='$id') AND ('$name' ='' OR Name LIKE '%$name%')"
Explanation:
First condition:
The query will select records with ID='$id' only when $id is not empty.
If $id is empty, query will not go for the second part ID='$id'
Second condition:
The query filters records with Name LIKE '%$name%' only when $name is not empty.
If $name is empty, query will not go for Name LIKE '%$name%'.
NB: This technique is extremely useful when you have numerous parameters to check, rather than using a bunch of if...elses at php side.
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
I have a column of pony names, where the breeder's Prefix is included with the pony's name (eg. Ashbrook Boy, where Ashbrook is the Breeder's Prefix, and Boy is the name of the pony). I have another table where I have a list of all the Prefixes used. I want to cycle through that list, and for each record search through my ponies and fetch those whose names begin with each prefix in turn. When they are fetched, I want to remove the said prefix from their name, and pop it into a column for that purpose on their own table.
In the end, I want -rather than one column with both Prefix and Name mixed in - two columns: one for Prefix, one for Name.
I thought the code below would do it for me, but it's not working. I get a 'not a valid resource' error for $res. Any help you could give me would be hugely appreciated - I really don't want to do this by hand! :P
I'm using a PHP script off a MySQL db, which I can access via PHPMyAdmin.
include '../conn.php';
$q=mysql_query("SELECT DISTINCT Pre FROM prefixes");
while($r=mysql_fetch_array($q)) {
$pre=$r['Pre'];
$sql="SELECT ID, Name FROM profiles WHERE (Name REGEXP '^$pre') ORDER BY ID ASC";
mysql_query($sql);
echo $sql;
while($res=mysql_fetch_array($sql)){
$name=$res['Name'];
$name=trim(str_replace("$pre","", $name));
$id=$res['ID'];
mysql_query("UPDATE profiles SET Prefix = '$pre', Name = '$name' WHERE ID = '$id' ");
}
}
mysql_close($con);
Your mistake come from the fact that $sql is the query string, not the mysql result of the query
$sql="SELECT ID, Name FROM profiles WHERE (Name REGEXP '^$pre') ORDER BY ID ASC";
mysql_query($sql);
echo $sql;
while($res=mysql_fetch_array($sql)){
with this, it will look better :
$sql="SELECT ID, Name FROM profiles WHERE (Name REGEXP '^$pre') ORDER BY ID ASC";
$result = mysql_query($sql);
while($res=mysql_fetch_array($result)){