I am using count function to get the number of rows buffered in a resultset.
But it always returns count as one even if resultset is empty.
Please see the code below:
$dbhandle = new SQLite3("sqlitedb_111.db");
$selQuery1 = "SELECT id,dbname,tabname,fieldname FROM scan_results ORDER BY id ASC LIMIT 0,10";
$resQuery1 = $dbhandle->query($selQuery1);
print count($resQuery1);
What am I doing wrong and how can I fix this?
As per your comment if you just want to return the count of records you could wrap your query in a SELECT COUNT(*) and change $dbhandle->query to $dbhandle->querySingle. This will work with or without LIMIT.
$dbhandle = new SQLite3("sqlitedb_111.db");
$selQuery1 = "SELECT COUNT(*) FROM (SELECT id,dbname,tabname,fieldname FROM scan_results ORDER BY id ASC LIMIT 0,10)";
$resQuery1 = $dbhandle->querySingle($selQuery1);
print count($resQuery1);
Result is an array. find the size of the array.
$dbhandle = new SQLite3("sqlitedb_111.db");
$selQuery1 = "SELECT id,dbname,tabname,fieldname FROM scan_results ORDER BY id ASC LIMIT 0,10";
$resQuery1 = $dbhandle->query($selQuery1);
$noofrows=sizeof($resQuery1);
echo $noofrows;
SQLite is an embedded database, i.e., there is no client/server communication overhead.
Therefore, it can return the results dynamically; there is only a single row buffered at any time.
To get the number of result rows, you either have to step through the results, or execute something like SELECT COUNT(*) FROM (original query).
Call numRows() on the result set.
From http://php.net/manual/en/function.sqlite-num-rows.php
<?php
$db = new SQLiteDatabase('mysqlitedb');
$result = $db->query("SELECT * FROM mytable WHERE name='John Doe'");
$rows = $result->numRows();
echo "Number of rows: $rows";
?>
Related
i have this code
$opslaglimit = 5;
$fag = "dansk";
$fagquery = "%".ucfirst($fag)."%";
$fagopslag = $db->prepare("
SELECT
*,
teacher_opslag.id as mainid,
teacher_opslag.created AS opslagcreated,
teacher_opslag.subject AS opslagsubject,
teacher_opslag.deleted AS maindeleted
FROM
teacher_opslag
WHERE
teacher_opslag.subject LIKE ?
ORDER BY
teacher_opslag.id DESC
LIMIT
$opslaglimit
");
$fagopslag->bind_param("s", $fagquery);
$fagopslag->execute();
$fagresult = $fagopslag->get_result();
$ensuranced = $fagresult->num_rows;
The query should select all data where dansk is present from the table. one of my rows might look like this Dansk_Engelsk_Svensk. The num_rows returns 0.
It is the first time i use LIKE in a WHERE clause. I dont know why it returns 0 when i have 3 matching rows in the db.
please help, thanks
I am using count function to get the number of rows buffered in a resultset.
But it always returns count as one even if resultset is empty.
Please see the code below:
$dbhandle = new SQLite3("sqlitedb_111.db");
$selQuery1 = "SELECT id,dbname,tabname,fieldname FROM scan_results ORDER BY id ASC LIMIT 0,10";
$resQuery1 = $dbhandle->query($selQuery1);
print count($resQuery1);
What am I doing wrong and how can I fix this?
As per your comment if you just want to return the count of records you could wrap your query in a SELECT COUNT(*) and change $dbhandle->query to $dbhandle->querySingle. This will work with or without LIMIT.
$dbhandle = new SQLite3("sqlitedb_111.db");
$selQuery1 = "SELECT COUNT(*) FROM (SELECT id,dbname,tabname,fieldname FROM scan_results ORDER BY id ASC LIMIT 0,10)";
$resQuery1 = $dbhandle->querySingle($selQuery1);
print count($resQuery1);
Result is an array. find the size of the array.
$dbhandle = new SQLite3("sqlitedb_111.db");
$selQuery1 = "SELECT id,dbname,tabname,fieldname FROM scan_results ORDER BY id ASC LIMIT 0,10";
$resQuery1 = $dbhandle->query($selQuery1);
$noofrows=sizeof($resQuery1);
echo $noofrows;
SQLite is an embedded database, i.e., there is no client/server communication overhead.
Therefore, it can return the results dynamically; there is only a single row buffered at any time.
To get the number of result rows, you either have to step through the results, or execute something like SELECT COUNT(*) FROM (original query).
Call numRows() on the result set.
From http://php.net/manual/en/function.sqlite-num-rows.php
<?php
$db = new SQLiteDatabase('mysqlitedb');
$result = $db->query("SELECT * FROM mytable WHERE name='John Doe'");
$rows = $result->numRows();
echo "Number of rows: $rows";
?>
I thought this would be simple but I'm having a tough time figuring out why this won't populate the the data array.
This simple query works fine:
$queryPrice = "SELECT price FROM price_chart ORDER BY id ASC LIMIT 50";
$resultPrice = mysqli_query($conn, $queryPrice);
$data = array();
while ($row = mysqli_fetch_array($resultPrice)) {
$data[] = $row[0];
}
But instead I want it to choose the last 10 results in Ascending order. I found on other SO questions to use a subquery but every example I try gives no output and no error ??
Tried the below, DOESN'T WORK:
$queryPrice = "SELECT * FROM (SELECT price FROM price_chart ORDER BY id DESC LIMIT 10) ORDER BY id ASC";
$resultPrice = mysqli_query($conn, $queryPrice);
$data = array();
while ($row = mysqli_fetch_array($resultPrice)) {
$data[] = $row[0];
}
I also tried specifying the table name again and using the IN, also doesn't work:
$queryPrice = "SELECT price FROM price_chart IN (SELECT price FROM price_chart ORDER BY id DESC LIMIT 10) ORDER BY id";
$resultPrice = mysqli_query($conn, $queryPrice);
$data = array();
while ($row = mysqli_fetch_array($resultPrice)) {
$data[] = $row[0];
}
In both examples my array is blank instead of returning the last 10 results and there are no errors, so I must be doing the subquery wrong and it is returning 0 rows.
The subquery doesn't select the id column, so you can't order by it in the outer query. Also, MySQL requires that you assign an alias when you use a subquery in a FROM or JOIN clause.
$queryPrice = "SELECT *
FROM (SELECT id, price
FROM price_chart
ORDER BY id DESC LIMIT 10
) x ORDER BY id ASC";
$resultPrice = mysqli_query($conn, $queryPrice) or die (mysqli_error($conn));
$data = array();
while ($row = mysqli_fetch_assoc($resultPrice)) {
$data[] = $row['price'];
}
You would have been notified of these errors if you called mysqli_error() when the query fails.
Your second query is the closest. However you need a table alias. (You would have seen this if you were kicking out errors in your sql. Note you will need to add any field that you wish to order by in your subquery. In this case it is id.
Try this:
SELECT * FROM (SELECT price, id
FROM price_chart ORDER BY id DESC LIMIT 10) as prices
ORDER BY id ASC
You must have errors, because your SQL queries are in fact incorrect.
First, how to tell you have errors:
$resultPrice = mysqli_query (whatever);
if ( !$resultprice ) echo mysqli_error($conn);
Second: subqueries in MySQL need aliases. So you need this:
SELECT * FROM (
SELECT id, price
FROM price_chart
ORDER BY id DESC LIMIT 10
) AS a
ORDER BY id ASC";
See the ) AS a? That's the table alias.
I am using the following query:
$query = "SELECT (SELECT count(*) FROM survey_ptw WHERE erfh= '0') AS F01,
(SELECT count(*) FROM survey_ptw WHERE erfh='10') AS F02,
(SELECT count(*) FROM survey_ptw WHERE erfh='50') AS F03";
$result = mysql_query($query);
print sprintf('Erf: <table><tr><td>none</td><td>%s</td></tr><tr><td>more</td><td>%s</td></tr></table>', $F01,$F02);
When I'm doing the above query in phpMyAdmin, it displays the variables nicely with the correct result values. Doing the same in PHP however, no results are displayed (empty strings)!
mysql_errno() and mysql_error() don't return errors. The DB has been open and selected further up in the code.
Any idea what's going on?
Because you aren't doing anything with the results.
From the documentation for mysql_query():
The returned result resource should be passed to mysql_fetch_array(), and other functions for dealing with result tables, to access the returned data.
So, to retrieve the results, you can use a while loop as below:
while ($row = mysql_fetch_assoc($result)) {
# code...
}
check $res=mysql_fetch_assoc($result); then print_r(); to check the result. it will work.
$query = "SELECT (SELECT count(*) FROM survey_ptw WHERE erfh= '0') AS F01,
(SELECT count(*) FROM survey_ptw WHERE erfh='10') AS F02,
(SELECT count(*) FROM survey_ptw WHERE erfh='50') AS F03";
$result = mysql_query($query);
$res=mysql_fetch_assoc($result);
$res=$res[0];
sprintf('Erf: <table><tr><td>none</td><td>%s</td></tr><tr><td>more</td><td>%s</td></tr></table>', $res['F01'],$res['F02']);
I have the selecting from the last ten entries working, but am unsure how to get the most popular from these ten entries? Also how would I count the number of the most popular entry & output it to a percentage?
<?php
$sql = "SELECT data FROM table_answers ORDER BY id DESC LIMIT 10";
$result = mysql_query ($sql, $db);
while ($row = mysql_fetch_array ($result))
{
echo "[".$row['data']."]";
}
?>
And I have tried to do the WHERE value as well but it doesn't return any result.
$sql = "SELECT data FROM table_answers WHERE id IN (SELECT id FROM table_answers
ORDER BY id DESC LIMIT 10) ORDER BY popularity DESC LIMIT 1";
$result = mysql_query ($sql, $db);
while ($row = mysql_fetch_array ($result))
{
echo " [".$row['data']."] ";
}
Anyone have any idea what I might be doing wrong here? please
This should solve the problem -
SELECT tableorder.*
FROM (SELECT *
FROM table
ORDER BY id DESC
LIMIT 10) tableorder
ORDER BY tableorder.popularity DESC
LIMIT 1
The inner query will sort on the basis on id and get the top 10. The outer will again sort the 10 rows on the basis of popularity and return the row with highest popularity.
SELECT data
FROM (
SELECT data
FROM table_answers
ORDER BY id DESC
LIMIT 10
) t
ORDER BY popularity