Following is my code,
$result1 = "SELECT emp_id FROM employee where manager_id=".$userID;
$array = mysql_query($result1);
$cnt = 0;
while ($row = mysql_fetch_array($array)) {
"emp_id: " . $row[0];
$myArrayOfemp_id[$cnt] = $row[0];
$cnt++;
}
var_dump($myArrayOfemp_id);
$sql = "SELECT emp_id FROM emp_leaves WHERE emp_id='$myArrayOfemp_id' ORDER BY apply_date DESC";
$result = mysql_query($sql);
$total_results = mysql_num_rows($result);
When I'am trying to use $myArrayOfemp_id variable in $sql query, It shows that error:
Array to string conversion in..
How can I fix it?
You are trying to convert an array into a string in the following line:
$sql = "SELECT emp_id FROM emp_leaves
WHERE emp_id='$myArrayOfemp_id' ORDER BY apply_date DESC";
$myArrayOfemp_id is an array. That previous line of code should be changed to:
$sql = "SELECT emp_id FROM emp_leaves
WHERE emp_id={$myArrayOfemp_id[0]} ORDER BY apply_date DESC";
I placed 0 inside {$myArrayOfemp_id[0]} because I'm not sure what value want to use that is inside the array.
Edited:
After discussing what the user wanted in the question, it seems the user wanted to use all the values inside the array in the sql statement, so here is a solution for that specific case:
$sql = "SELECT emp_id FROM emp_leaves
WHERE ";
foreach ($myArrayOfemp_id as $value)
{
$sql .= " emp_id={$value) || ";
}
$sql .= "1=2";
$result = mysql_query($sql);
$total_results = mysql_num_rows($result);
$sql = "SELECT emp_id FROM emp_leaves WHERE emp_id in
(SELECT GROUP_CONCAT(emp_id) FROM employee where manager_id=".$userID.")
ORDER BY apply_date DESC";
$result = mysql_query($sql);
$total_results = mysql_num_rows($result);
just change your query like above might solve your problem.
you can remove following code now. :)
$result1 = "SELECT emp_id FROM employee where manager_id=".$userID;
$array = mysql_query($result1);
$cnt = 0;
while ($row = mysql_fetch_array($array)) {
"emp_id: " . $row[0];
$myArrayOfemp_id[$cnt] = $row[0];
$cnt++;
}
var_dump($myArrayOfemp_id);
Related
i'm trying to sum a column name "total". and i want to display the total sorting by id. if user A login he can see total booking in his account.
I keep get the error:
"Notice: Array to string conversion in Array."
can someone help me? I want to echo the total in input form.
this is my php code:
<?php
include ('connect.php');
$sql = "SELECT * FROM penjaga WHERE p_username = '".$_SESSION['username']."'";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_assoc($result);
$id = $row['p_id'];
$sql2 = "SELECT SUM(total) as total FROM sitter_kucing WHERE sitter_fk = '$id'";
$row2 = mysql_fetch_array($sql);
$sum = $row['total'];
?>
Try this,
$sql2 = "SELECT SUM(total) as total FROM sitter_kucing WHERE sitter_fk = '$id'";
$result2 = mysql_query($sql2) or die(mysql_error());
$row2 = mysql_fetch_array($result2) or die(mysql_error());
$sum = $row['total'];
i got it! thanks this is my code
this is the code:
<?php
include ('connect.php');
$sql8 = "SELECT * FROM penjaga WHERE p_username = '".$_SESSION['username']."'";
$result8 = mysqli_query($conn,$sql8);
$row8 = mysqli_fetch_assoc($result8);
$id = $row8['p_id'];
$sql9 = "SELECT SUM(total) as total FROM sitter_kucing WHERE sitter_fk = '$id'";
$result9 = mysqli_query($conn,$sql9);
$row9 = mysqli_fetch_array($result9);
$sum = $row9['total'];
?>
I am trying to get the key value from the multidimensinal array which I have created using .The Array snapshot is given after the Code.
Below is my PHP code-
$selectTicket = "select ticketID from ticketusermapping where userID=$userID and distanceofticket <=$miles;";
$rsTicket = mysqli_query($link,$selectTicket);
$numOfTicket = mysqli_num_rows($rsTicket);
if($numOfTicket > 0){
$allRowData = array();
while($row = mysqli_fetch_assoc($rsTicket)){
$allRowData[] = $row;
}
$key = 'array(1)[ticketID]';
$QueryStr = "SELECT * FROM ticket WHERE ticketID IN (".implode(',', array_keys($key)).")";
Array Snapshot-
I need the tickedID value from this array . Like the first one is 49 .
Please help.
change your code like
$selectTicket = "select ticketID from ticketusermapping where userID=$userID and distanceofticket <=$miles;";
$rsTicket = mysqli_query($link, $selectTicket);
$numOfTicket = mysqli_num_rows($rsTicket);
if ($numOfTicket > 0) {
$allRowData = array();
while ($row = mysqli_fetch_assoc($rsTicket)) {
$allRowData[] = $row['ticketID'];
}
$QueryStr = "SELECT * FROM ticket WHERE ticketID IN (" . implode(',', $allRowData) . ")";
$ids = array_column( $allRowData, 'ticketID'); //this will take all ids as new array
$QueryStr = "SELECT * FROM ticket WHERE ticketID IN (".implode(',', $ids).")";
You should do a single query using JOIN for this:
$query = "
SELECT t.*
FROM ticket t
JOIN ticketusermapping tum
ON t.ticketID = tum.ticketID
AND tum.userID = '$userID'
AND tum.distanceofticket <= '$miles'
";
$stmt = mysqli_query($link, $query);
$numOfTickets = mysqli_num_rows($stmt);
while($row = mysqli_fetch_assoc($stmt)){
var_dump($row); // here will be the ticket data
}
Not a duplicate of Select specific value from a fetched array
I have a MySql database as:
Here's my query:
$sql = "SELECT * FROM data ORDER BY Score DESC";
I want it to be a leaderboard which people can update their scores so I can't use
$sql = "SELECT * FROM data ORDER BY Score DESC WHERE ID = 1";
I want to get Username of the second row in my query.So I wrote:
<?php
include "l_connection.php";
$sql = "SELECT * FROM data ORDER BY Score";
$result = mysqli_query($conn, $sql);
if($result->num_rows>0){
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
}
echo "Result = '".$row[1]['Username']."''";
}
?>
But it returns Result = '' like there's nothing in the array.
But if I write
if($result->num_rows>0){
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "Name = '".$row['Username']."''";
}
}
It will return : Parham, Mojtaba, Gomnam, Masoud,
So what am I doing wrong in the first snippet?
You can not access $row outside of while loop.
So store result in one new array, and then you can access that new array outside the while loop:
$newResult = array();
if($result->num_rows>0){
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$newResult[] = $row;
}
}
echo "Result = '".$newResult[1]['Username']."''"; // thus you can access second name from array
Because you write where condition after ORDER by at
$sql = "SELECT * FROM data ORDER BY Score DESC WHERE ID = 1";
The sequence of query is
SELECT * FROM data // select first
WHERE ID = 1
ORDER BY Score DESC// Order by at last
Check http://dev.mysql.com/doc/refman/5.7/en/select.html
And for second case you need to fetch Username inside while loop and use $row['Username'] instead $row[1]['Username']
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "Result = '".$row['Username']."''";// write inside while loop
}
You can assign row value to any array and use that array.
<?php
include "l_connection.php";
$sql = "SELECT * FROM data ORDER BY Score";
$result = mysqli_query($conn, $sql);
if($result->num_rows>0){
$rows = array();
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$rows[] = $row;
}
echo "Result = '".$rows[1]['Username']."'";
}
?>
Or if you want only second highest score from column you can user limit as
$sql = "SELECT * FROM data ORDER BY Score limit 1,1";
I have a query that gets 5 lines of data like this example below
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
}
I want to run a query inside each results like this below
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}
but for some reason it's only display the first line when I add the query inside it. I can seem to make it run all 5 queries.
SELECT
t1.ref,
t1.user,
t1.id,
t2.domain,
t2.title
FROM
table AS t1
LEFT JOIN anothertable AS t2 ON
t2.domain = t1.ref
LIMIT
0, 5
The problem is that inside the while-cycle you use the same variable $result, which then gets overridden. Use another variable name for the $result in the while cycle.
You change the value of your $query in your while loop.
Change the variable name to something different.
Ex:
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$qry = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$rslt = mysql_query($qry) or die(mysql_error());
if (mysql_num_rows($rslt) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}
Use the following :
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query_domain = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result_domain = mysql_query($query_domain) or die(mysql_error());
if (mysql_num_rows($result_domain) )
{
$row_domain = mysql_fetch_row($result_domain);
$title = $row_domain['title'];
} else {
$title = "No Title";
}
echo "$ref - $title";
}
This is a logical problem. It happens that way, because you are same variable names outside and inside the loop.
Explanation:
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
// Now $results hold the result of the first query
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
//Using same $query does not affect that much
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
//But here you are overriding the previous result set of first query with a new result set
$result = mysql_query($query) or die(mysql_error());
//^ Due to this, next time the loop continues, the $result on whose basis it would loop will already be modified
//..............
Solution 1:
Avoid using same variable names for inner result set
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$sub_result = mysql_query($query) or die(mysql_error());
// ^ Change this variable so that it does not overrides previous result set
Solution 2:
Avoid the double query situation. Use joins to get the data in one query call. (Note: You should always try to optimize your query so that you will minimize the number of your queries on the server.)
SELECT
ref,user,id
FROM
table t
INNER JOIN
anothertable t2 on t.ref t2.domain
LIMIT 0, 5
Learn about SQL joins:
SELECT table.ref, table.user, table.id, anothertable.title
FROM table LEFT JOIN anothertable ON anothertable.domain = table.ref
LIMIT 5
You're changing the value of $result in your loop. Change your second query to use a different variable.
it is not give proper result because you have used same name twice, use different name like this edit.
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query1 = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result1 = mysql_query($query1) or die(mysql_error());
if (mysql_num_rows($result1) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}
I am trying to store a mysql value into a php variable. I have the following query which I know works. However, I the value for $count is always 0. Can someone explain what I need to do to get the count value? The count should be the count of x's w here name_x=.$id.
$query = "SELECT COUNT(name_x) FROM Status where name_x=.$id.";
$result = mysql_query($query);
$count = $result;
Is first letter in table name is really capital. Please check it first.
or Try :
$query = "SELECT COUNT(*) as totalno FROM Status where name_x=".$id;
$result = mysql_query($query);
while($data=mysql_fetch_array($result)){
$count = $data['totalno'];
}
echo $count;
$query = "SELECT COUNT(*) FROM `Status` where `name_x`= $id";
$result = mysql_query($query);
$row = mysql_fetch_row($result);
$count = $row[0];
please try it
$query = "SELECT COUNT(*) FROM Status where name_x=$id";
$result = mysql_query($query);
$count = mysql_result($result, 0);
You are missing single quotes around $id. Should be
name_x = '" . $id . "'";