I am learning JSON.
getDatas.php contents:
$query = "SELECT domain FROM access WHERE userid = '".$userid."'";
$result = mysql_query($query) or die(mysql_error());
$res = array();
while($row = mysql_fetch_array($result)){
$query2 = "SELECT COUNT(*) as cnt from clients WHERE domain ='".$row['domain']."'";
$res2 = mysql_query($query2) or die(mysql_error());
$res2 = mysql_fetch_array($res2);
$res[] = array('domain' => $row['domain'], 'count' => $res2['cnt']);
};
echo json_encode($res);
The output is:
[{"domain":"www.domain1.com","count":"2"},{"domain":"www.domain2.com","count":"42"},{"domain":"www.domain3.com","count":"61"}]
How do I print like this?
How do I get out clean?
www.domain1.com - 2
www.domain2.com - 42
www.domain3.com - 61
Try this:
$query = "SELECT a.domain, COUNT(c.domain) as cnt
FROM `access` as a
LEFT JOIN `clients` as c on c.domain = a.domain
WHERE a.userid = '".$userid."'
GROUP BY c.domain";
$result = mysql_query($query) or die(mysql_error());
$res = array();
while($row = mysql_fetch_array($result)){
$res[] = array($row['domain'] => $row['cnt']);
};
echo json_encode($res);
you only need to use
echo json_encode($res);
Related
I have case manager table where i have inserted court table id as foreign key. i want to fetch record from both tables. when using nested while loop it shows only one row data.
$id = $_SESSION['id'];
$query1 = "SELECT * from `case_manager` where user_id = '$id' ";
$result1 = mysqli_query($conn, "$query1");
while($row = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {
$Status = $row['status'];
$id = $row['id'];
$case_type = $row['case_type'];
$court_id = $row['court_id'];
$query2 = "SELECT * from `case_type` where case_id = '$case_type'";
if($result1 = mysqli_query($conn, "$query2")) {
while($row2 = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {
echo $row2['case_name'];
}
}
}
Because you are overwritting you $result1 change inner query result to $result2 then try
$id = $_SESSION['id'];
$query1 ="SELECT * from `case_manager` where user_id = '$id' ";
$result1 = mysqli_query($conn , "$query1");
while ($row = mysqli_fetch_array($result1 ,MYSQLI_ASSOC)) {
$Status=$row['status'];
$id = $row['id'];
$case_type = $row['case_type'];
$court_id = $row['court_id'];
$query2 ="SELECT * from `case_type` where case_id = '$case_type'";
if($result2 = mysqli_query($conn , "$query2")){;
while ($row2 = mysqli_fetch_array($result2 ,MYSQLI_ASSOC)) {
echo $row2['case_name'];
}
}
}
1st : Because you are overwriting the variable $result1 In second query execution.
if($result1 = mysqli_query($conn , "$query2")){;
^^^^^^^^ ^^
Note : And remove that unnecessary semicolon .
2nd : No need multiple query simple use join
SELECT cm.*,c.* from `case_manager` cm
join `case_type` c
on cm.cas_type=c.case_id
where cm.user_id=$id;
You can use below query to fetch your record:
$query = SELECT case_manager.* ,case_type.case_name FROM case_manager Left JOIN case_type ON case_manager.case_type=case_type.case_id where case_manger.user_id = $id;
While($row = mysql_fetch_array()){
echo $row['case_name'];
}
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
}
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);
This is my code and I don't know why IF/ELSE statement is not working
$user_id = $_POST['user_id'];
$strSQL = "SELECT chat.user_id,max(date) max_date ,user.user_id,user.firstname,user.lastname,user.picture
FROM ( select from_user_id as user_id,date,isread from chat
WHERE to_user_id = '$user_id'
Union select to_user_id as user_id,date,isread from chat WHERE from_user_id = '$user_id' ) as chat join user on user.user_id = chat.user_id
where user.status != 'block' group by chat.user_id order by max_date DESC";
$chatdata = array();
$objQuery = mysql_query($strSQL);
while($row = mysql_fetch_assoc($objQuery)){
$frduser_id = $row['user_id'];
$firstname = $row['firstname'];
$lastname = $row['lastname'];
$picture = $row['picture'];
$strSQL2 = "SELECT isread from chat WHERE to_user_id = '$user_id' and from_user_id = '$frduser_id'
ORDER BY date DESC LIMIT 1";
//array_push($chatdata, $row);
$objQuery2 = mysql_query($strSQL2);
if (empty($objQuery2)) {
$row2 = 1;
$result = array_merge($row, $row2);
array_push($chatdata,$result);
}else{
$row2 = mysql_fetch_assoc($objQuery2);
$result = array_merge($row, $row2);
array_push($chatdata,$result);
}
}
You can use mysql_num_rows to get count of rows inside results.
$objQuery2 = mysql_query($strSQL2);
$objQuery2Rows = mysql_num_rows($strSQL2);
if ( $objQuery2Rows < 1) {
$row2 = 1;
$result = array_merge($row, $row2);
array_push($chatdata,$result);
}else{
$row2 = mysql_fetch_assoc($objQuery2);
$result = array_merge($row, $row2);
array_push($chatdata,$result);
}
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";
}