Happy holidays, if you celebrate it.
I am trying to run the following query:
$query = SELECT WHERE EXISTS (SELECT * FROM table WHERE val1="someVal" and val2="someVal");
$result = mysql_query ($query)
print_r (mysql_fetch_assoc($result));
My goal is to create an associative array with only records from the matching row that have a value.
Am I using EXISTS in the wrong manner?
h2ooooooo is right, you don't need to use exits to select what you want here. Just a simple WHERE clause will do it. also I would single quote the values not double quote them for mysql syntax purposes. ie:
$query = "SELECT * FROM table WHERE val1='someVal' and val2='someVal'";
$result = mysql_query($query, $link);
print_r (mysql_fetch_assoc($result));
Related
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.
I know this is very simple, but I haven't used PHP/MySQL in a while and I have been reading other threads/php website and can't seem to get it.
How can I query a single row from a MySQL Table and print out all of the fields that have data in them? I need to exclude the NULL fields, and only add those that have data to an html list.
To clarify, I would like to display the field data without specifying the field names, just for the reason that I have a lot of fields and will not know which ones will be NULL or not.
What you've outlined requires 4 basic steps:
Connect to the database.
Query for a specific row.
Remove the null values from the result.
Create the html.
Step 1 is quite environment specific, so that we can safely skip here.
Step 2 - SQL
SELECT * from <tablename> WHERE <condition isolating single row>
Step 3 - PHP (assuming that $query represents the executed db query)
//convert the result to an array
$result_array = mysql_fetch_array($query);
//remove null values from the result array
$result_array = array_filter($result_array, 'strlen');
Step 4 - PHP
foreach ($result_array as $key => $value)
{
echo $value \n;
}
Just SELECT * FROM table_name WHERE.... will do the trick.
To grab data from specific fields, it would be SELECT field_1,field_2,field_3....
you have to make a string which represent mysql query. Then there is function in php named mysql_query(). Call this function with above string as parameter. It will return you all results. Here are some examples
You need to do it like this...
First connect to your sql... Reference
Now make a query and assign it to a variable...
$query = mysqli_query($connect, "SELECT column_name1, column_name2 FROM tablename");
If you want to retrieve a single row use LIMIT 1
$query = mysqli_query($connect, "SELECT column_name1, column_name2 FROM tablename LIMIT 1");
If you want to fetch all the columns just use * instead of column names and if you want to leave some rows where specific column data is blank you can do it like this
$query = mysqli_query($connect, "SELECT * FROM tablename WHERE column_name4 !=''");
Now fetch the array out of it and loop through the array like this..
while($show_rows = mysqli_fetch_array($query)) {
echo $show_rows['column_name1'];
echo $show_rows['column_name2'];
}
If you don't want to include the column names in the while loop, you could do this:
while($show_rows = mysqli_fetch_array($query)) {
foreach( $show_rows as $key => $val )
{
echo $show_rows[$key];
}
}
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`.
Im trying to pass an array that I already found by a query into another query. For example:
$first_query = "SELECT id FROM from Table WHERE user = '{$_SESSION['id'}'";
$result = mysql_query($first_query,$connection);
$ids = mysql_fetch_array($result);
This is where it gets tricky for me. I want to pass $ids into the next query.
$id_implode = implode(", ", $ids)
$second_query = "SELECT * FROM Table2 WHERE id = '{$id_implode}';
The second query doesnt seem to be working. Any help is greatly appreciated!
your second query's syntax is wrong. Once evaluated it should read
select * from Table2 where id in (1,2,3)
ditch the curly braces and change the = to in. Don't use OR - that's a dumb way of ignoring good sql functionality
EDIT: Teneff's comment makes a very good point - why are you approaching the problem in this way? If there is a relationship between the tables they can be joined and all the data you want can be retrieved in a single query. If for some reason you can't / won't join the tables you could at least try a sub-query
select * from table2 where id in (select id from table where user = $_SESSION['id']);
To use a where statement with multiple entries to match on, use in ().
$id_implode = "'".implode("', '", $ids)."'"
$second_query = "SELECT * FROM Table2 WHERE id in ({$id_implode});
I think you should use IN
$id_implode = implode(", ", $ids)
$second_query = "SELECT * FROM Table2 WHERE id IN '({$id_implode})';
This assumes that $ids is made of int of course, otherwise you have to enclose eache entry in quotes. that means
IN (6,7,8,9)//this doesn't need quotes
IN ('lemon', 'orange')//needs quotes
try to use the IN syntax:
$id_implode = implode("', '", $ids);
$second_query = "SELECT * FROM Table2 WHERE id in ('{$id_implode}');
// make empty array
$sqlArray=array();
$jsonArray=array();
// START NEED FAST WORKING ALTERNATİVES -----------------------------------------------------
// first 20 vistors
$query = "SELECT user_id FROM vistors LIMIT 20";
$result = mysql_query ($query) or die ($query);
// make vistors user query array
while ($vstr_line = mysql_fetch_array($result)){
array_push($sqlArray, $vstr_line['user_id']);
}
// implode vistors user array
$sqlArray_impl = implode("', '", $sqlArray);
// END NEED FAST WORKING ALTERNATİVES -----------------------------------------------------
// Get vistors information
$query = "SELECT id, username, picture FROM users WHERE id IN ('$sqlArray_impl')";
$qry_result = mysql_query($query) or die($query);
while ($usr_line = mysql_fetch_array($qry_result)){
array_push($jsonArray, $usr_line['id'].' - '.$usr_line['username'].' - '.$usr_line['picture']);
}
print_r($sqlArray);
echo '<br><br>';
print_r($jsonArray);
see this my functions..
i need a replacement for fast working alternatives..
function within the range specified above, to me, running faster than the alternative.
the query will return back array ?
thx for all helpers !
Can you use a JOIN or SUB SELECT to reduce the query count from 2 to 1? Might not give much of a boost but worth a shot and a cleaner implementation.
Where is the bottleneck? Most likely the db and not the php code.
Are the tables/columns properly indexed? Run EXPLAIN on both queries.
Easiest would be to include first query as subquery eliminating one turn to the DB and a lot of code:
// Get vistors information
$query = "SELECT id, username, picture FROM users WHERE id IN (SELECT user_id FROM vistors LIMIT 20)";
$qry_result = mysql_query($query) or die($query);
Unless there is more reason to have the first one seperate, but that is not visible in your code example.
If you use PDO (recommended anyway...), you can return the result array all at once using fetchAll().
For your second query, you can use string concatenation in MySQL to directly return the result you want.