I have this statement that's working well in SQL:
SET #cnt = (SELECT `reg_date` FROM logindb WHERE `userlogin`='urs');
SET #var = (SELECT TIME (#cnt));
SELECT #var
But I don't know how to get that data if I don't have a specific column output.
In php I'm trying with the following code:
$sql = "SET #cnt = (SELECT `reg_date` FROM logindb WHERE `userlogin`='urs');
SET #var = (SELECT TIME (#cnt));
SELECT #var"
$result = $conn->query($sql);
$row = $result->fetch_assoc();
echo($row["#var"]);
Fatal error: Call to a member function fetch_assoc() on boolean in C:\xampp\htdocs\DAN\autentificare.php on line 62
The line 62 :
$row = $result->fetch_assoc();
I don't know how to store or view that variable in PHP.
Explanation of the code:
When I created the table named logindb, I use a variable reg_date declared as TIMESTAMP that put the server time every time another column from the same line is filled. To resume the entire code is used to get only the TIME from reg_date : Ex. 2015-01-06 00:39:56 output that I get in #var is 00:39:56
Try this:
SELECT #var as colName
....
echo($row["colName"]);
UPDATE
When I read the post first time I think that the question is something simplified. Anyway, to solve your issue you can use this code.
$sql = "SELECT TIME(`reg_date`) FROM logindb WHERE `userlogin`='urs'";
$result = $conn->query($sql);
echo $result->fetch_row[0];
The following code(assuming your $sql statement is valid)
$row = $result->fetch_assoc();
Returns an associative array. To access the value the key would look likethis:
$myresult = $row['column'];
*column is the column you selected in your statement, in your case 'reg_date'
Related
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,$query2);
}
}
I know how to select from a database, my question is i want to set a variable for a where clause in my query, so lets say select * FROM Foo WHERE A = '$a', but the variable has not been set previously and is from the table i am trying to select from.
So how would i do this.
PHP
$FC_ID = mysqli_real_escape_string($dbc,[]);
$FOOD_sel = $dbc->query("SELECT * FROM Food_Cat WHERE Food_Cat_ID = '$FC_ID'");
That is how far i have got so far.
Try to use sub query for that $FC_ID value
sub query method
$FC_ID = mysqli_real_escape_string($dbc,[]);
$FOOD_sel = $dbc->query("SELECT * FROM Food_Cat WHERE Food_Cat_ID = (select FC_ID from your table )");
The query SELECT * FROM TABLE WHERE id LIKE '%1% is not working properly, it's not select the id 1.
mysql_connect('localhost', 'root' , '');
mysql_select_db('database');
$sql = ("select * from search WHERE id LIKE '%3%'");
mysql_query($sql);
$my_variable = mysql_query($sql);
$display_data = mysql_fetch_row($my_variable);
while ($list = mysql_fetch_assoc($my_variable)) {
$id = $list['id'];
$title = $list['title'];
$keywords = $list['keywords'];
$img = $list['img'];
$link = $list['link'];
}
If you are looking to SELECT id 1 then use = not LIKE. The way LIKE is being used it will match every id that has a 1 in it and you are not guaranteed to get the first one in order, so instead use:
SELECT * FROM search WHERE id = 1
According to the PHP documentation of mysql_fetch_row it
Returns a numerical array that corresponds to the fetched row and moves the internal data pointer ahead.
Which means that the first result won't show up in the next (mysql_fetch_assoc) procedure. You could try removing the $display_data = mysql_fetch_row($my_variable); line and only use the while($list = mysql_fetch_assoc($my_variable)) { ... } procedure. See if that solves your problem.
$sql = ("select * from search WHERE id ='3'");
The id is an integer use = instead of like . Equal is more accurate.
And first echo your query in your program->
echo $sql;die;
copy that query and run it on your phpmyadmin
and then check is your column id is int type if it is then like will not give you the result. You have to use the where clause here .But if you have the column id is of type varchar then definitely give you the result .
Try to use search tab under your database->table in your phpmyadmin and put the condition there.
You will definitely get your answer there.
I've been using this command to retrieve the number of the fields which have same email address:
$query = $db->query("SELECT COUNT(`user_id`) FROM `users` WHERE `email`='$email'") or die($db-error);
There are 3 records in users table with the same email address. The problem is when I put * instead of COUNT(user_id) it returns correctly: $query->num_rows gives 3 but when I use COUNT(user_id) then $query->num_rows returns 1 all the time. how can I correct this or where is my problem?
When you use $query->num_rows with that query it will return 1 row only, because there is only one count to return.
The actual number of rows will be contained in that query. If you want the result as an object, or associative array give the count a name:
$query = $db->query("SELECT COUNT(`user_id`) AS total FROM `users` WHERE `email`='$email'") or die($db-error);
And in the returned query total should be 3, while $query->num_rows will still be 1. If you just want the value a quick way would be using $total = $query->fetchColumn();.
As others have said though, be careful with NULL user ids, because COUNT() will ignore them.
Emails have to be uinque in users table. Thus, you need no count at all.
You ought to use prepared statements.
You shouldn't post a code that will never run.
Here goes the only correct way to run such a query:
$sql = "SELECT * FROM `users` WHERE `email`=?";
$stm = $db->prepare($sql);
$stm->execute([$email]);
$user = $stm-fetch();
(the code was written due to erroneous tagging. For mysqli you will need another code, but guidelines remains the same.)
Something like this
$sql = "SELECT * FROM `users` WHERE `email`=?";
$stm = $db->prepare($sql);
$stm->bind_param('s',$email);
$stm->execute();
$res = $stm->get_result()
$user = $res->fetch_assoc();
in $user variable you will have either userdata you will need in the following code or false which means no user found. Thus $user can be used in if() statement all right without the need of any counts.
In case when you really need to count the rows, then you use this count() approach you tried. You can use a function from this answer for this:
$count = getVar("SELECT COUNT(1) FROM users WHERE salary > ?", $salary);
That's the correct behaviour: If you use the COUNT function, the result of your select query will be just one row with one column containing the number of data sets.
So, you can retrieve the number of users with the given E-mail address like this:
$query = $db->query("SELECT COUNT(`user_id`) FROM `users` WHERE `email`='$email'") or die($db-error);
$row = $query->fetch_row();
$count = $row[0];
Note that this is faster than querying all data using SELECT * and checking $query->num_rows because it does not need to actually fetch the data.
For the user I am testing with, their org_id column value is "student_life"
I am trying to have this function display whatever rows have the student_life column = 1. (so yes there is a column student_life which is a boolean, and then I also have a separate column named org_id and in this case has the value student_life)
I am pretty sure there is a syntax error but I cannot figure it out.
function org_id_users_table()
{
$org_id = mysql_real_escape_string($_POST["org_id"]);
$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE '$org_id' = '1'");
$result = $sql['sql'];
$num_rows = $sql['num_rows'];
$this->create_table($result, $num_rows);
}
(when I replace $org_id in the "$sql=..." line with student_life the code works.
You're quoting the column name, which makes MySQL think it's a string.
$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE $org_id = '1'");
Edit:
Based on your comments, I think what you actually want is this:
$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE org_id = '$org_id'");
Change quotes.
$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE `$org_id` = '1'");
P.S. Why shouldn't I use mysql_* functions in PHP?
Where is this coming from? $_POST["org_id"]
Do you have a form on the page posting that? Or are you just trying to get that from the database? If so, wouldn't you need another query to obtain that first?
$row_MyFirstQuery['org_id']
Otherwise if it is $_POST["org_id"], wouldn't it be single quotes not double? $_POST['org_id']