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'];
?>
Related
I want to show the count of users which have the status 1 (see code) within PHP MySQL.
<?php
// something like this
$result = mysqli_query($mysqli, "SELECT COUNT(*) FROM users WHERE status = '1'");
echo "$result";
?>
Try this:
$query = "SELECT COUNT(*) as countvar FROM users where status = '1'";
$result = mysqli_query($con,$query);
$row = mysqli_fetch_array($result);
$count= $row['countvar '];
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'];
}
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);
Im trying to generate an array but not sure how to go about it.
I'm currently getting my data like so:
$query = mysql_query("SELECT * FROM users WHERE userEmail LIKE 'test#test.com'");
$row = mysql_fetch_array($query);
$query1 = mysql_query("SELECT * FROM categories");
while($row1 = mysql_fetch_array($query1)){
$query2 = mysql_query("SELECT * FROM usersettings WHERE userId = ".$row['userId']." AND usersettingCategory".$row1['categoryId']." LIKE 'y'");
$isyes = mysql_num_rows($query2);
if($isyes > 0){
$cat1 = mysql_query("SELECT * FROM shops WHERE shopstateId = 1 AND (categoryId1 = ".$row1['categoryId']." OR categoryId2 = ".$row1['categoryId']." OR categoryId3 = ".$row1['categoryId'].")");
$cat1match = mysql_num_rows($cat1);
if($cat1match > 0){
while($cat1shop = mysql_fetch_array($cat1)){
$cat1msg = mysql_query("SELECT * FROM messages WHERE shopId = ".$cat1shop['shopId']." and messagestateId = 1");
while($cat1msgrow = mysql_fetch_array($cat1msg)){
echo $cat1msgrow['messageContent']." - ".$cat1msgrow['messageCode'];
$cat1img = mysql_query("SELECT shopimagePath FROM shopimages WHERE shopimageId = ".$cat1shop['shopimageId']);
$imgpath = mysql_fetch_array($cat1img);
echo " - ".$imgpath['shopimagePath']."<br/>";
}
}
}
}
}
But this can cause duplicates when a user has all 3 of a shops categories picked in their preferences. I am trying to find a way to just pull the message ID out instead of the whole thing and put it into an array giving me, for example:
1,3,5,7,1,3,5,2,4,7,8
Then I can just run a separate query to say get me all messages where the ID is in the array, but i am unsure of the most constructive way to build such an array and examples of array from a while loop I have seen do not seem to be what I am looking for.
Is there anyone out there that can push me in the right direction?
Can't help with this code. But if you want an array from a query without duplicate result, you can use " select DISTINCT (id) " in your query or for more simple solution :
$id_arr = array();
$sql = mysql_query("select id from id_table");
while ($id_result = mysql_fetch_array($sql) {
$id = $id_result['id'];
if (!in_array($id, $id_arr)) {
$id_arr[] = $id;
}
}
I have found a much easier way to create the required result. I think at 6am after a hard night coding my brain was fried and I was making things a lot more complicated than I needed to. A simple solution to my issue is as follows:
$query = mysql_query("SELECT * FROM users WHERE userEmail LIKE 'test2#test2.com'");
$row = mysql_fetch_array($query);
$categories = "(";
$query1 = mysql_query("SELECT * FROM categories");
while($row1 = mysql_fetch_array($query1)){
$query2 = mysql_query("SELECT usersettingCategory".$row1['categoryId']." FROM usersettings WHERE userId = ".$row['userId']);
$row2 = mysql_fetch_array($query2);
if($row2['usersettingCategory'.$row1['categoryId']] == y){
$categories .= $row1['categoryId'].",";
}
}
$categories = substr_replace($categories ,")",-1);
echo $categories."<br />";
$query3 = mysql_query("SELECT * FROM shops,messages WHERE shops.shopId = messages.shopId AND messages.messagestateId = 1 AND (shops.categoryId1 IN $categories OR shops.categoryId2 IN $categories OR shops.categoryId3 IN $categories)");
while($row3 = mysql_fetch_array($query3)){
$query4 = mysql_query("SELECT shopimagePath FROM shopimages WHERE shopimageId = ".$row3['shopimageId']);
$row4 = mysql_fetch_array($query4);
echo $row3['messageContent']." - ".$row3['messageCode']." - ".$row4['shopimagePath']."<br />";
}
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 . "'";