I'm running a query to get the total sum of some hours from a database table to show in a PDF file but for some strange reason it won't echo 0 (zero).
$query = mysql_query("SELECT id, SUM(hours_night), SUM(hours_days) FROM table WHERE invoiceID='".mysql_real_escape_string($invoiceID)."'") or die(mysql_error());
$result = mysql_num_rows($query);
$totalhours_night = 0;
while ($fetch = mysql_fetch_assoc($query)) {
$totalhours_night += $fetch['SUM(hours_night)'];
}
$html_output = "Some html and tables markup... " . $totalhours_night . "";
The output of the html works fine, thats not the problem.. but the problem is it won't output 0 (zero) for some reason. If the hour result is actualy something like 1 or 5 or whatever it outputs the totalhours fine, but i need it to output 0 if there are no hours.
Because it looks strange to ouput nothing if there are no hours, i need to display a 0 zero since this looks more good.
BTW, if i for example put number_format($totalhours_night, 2); it does display 0.00, but i need it to be just 0.
Three things:
Your SQL query is wrong, you can't display id and SUM(..) unless you use GROUP BY id (Which kid of loses the idea if id is unique)
If you use SUM(..) give it an alias like SUM(..) AS alias1, and later in the PHP refer to alias1.
If this is not the reason, try var_dump($fetch['alias1']); and look at what you really get back. (And post your findings here so we can give you further help)
Treat integer as string like this-
echo (string)$totalhours_night;
Related
I am using PHP lite to search for a matching row like this...
$count = $db->exec("SELECT * FROM users WHERE userid = '34534fgr'");
echo $count;
But my count is returning 1 every time, even when the value does not exist.
Am I searching incorrectly?
Am I searching incorrectly?
yup. protip: when debugging, use var_dump instead of echo, it would help you see the issue here, because it would print bool(true); instead of int(1) or string("1"), because PDO::exec() returns a boolean.
here's how to do what you tried to do:
$count = $db->query("SELECT COUNT(*) FROM users WHERE userid = '34534fgr'",PDO::FETCH_NUM)->fetch()[0];
I'm new to PHP and i want to know how i can subtract a specific amount from the results from counting the total amount of rows in a table. In this case i'd like to minus the value 3 from whatever the value of the total rows is. But i keep getting an error. Below is my code.
$cartwork = $con->query("SELECT count(*) FROM table");
$vs = '3';
$camount = $cartwork - $vs;
echo "$camount";
When the code runs i get the error "Object of class mysqli_result could not be converted to int" what can i do to fix this and get it to work properly.
The query returns a result set. You need to parse through the result set(s) in order to access the values returned. That's basically what the error states.
Please see here for documentation on the PHP function for fetching rows:
http://php.net/manual/en/function.mysql-fetch-row.php
So basically you would need
$row=$cartwork->mysql_fetch_row();
$cartWork_value = $row[0];
$vs = '3';
$camount = $cartwork_Value - $vs;
echo "$camount";
Note - this assumes that you get back exactly one result row (which should be the case with your query).
You can simply change your query to:
$cartwork = $con->query("SELECT count(*)-3 FROM table");
It doesn't smell particularly good though.
I'm trying to run if and else if inside a while loop in my PHP code.
The code looks like this:
<?php
$sql = "SELECT * FROM table ORDER BY id";
$query = mysqli_query($db_conx, $sql);
$productCount = mysqli_num_rows($query);
if ($productCount > 0) {
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
$deviceType = $row["deviceType"];
if($deviceType == 'iPhone' || $deviceType == 'iPad'){
echo 'IOS';
}else if($deviceType == 'Android'){
echo 'Android';
}
}
} else {
}
?>
The code above works (sort of) but not as I was expecting it.
To give you an example, lets say I have 2 rows in MYSQL database.
like this:
id deviceType
1 Android
2 iPhone
when i run my PHP code above, I get this echo-ed on my page:
IOS
IOS
Android
Android
BUT I only have 2 rows in the database so the result should be:
IOS
Android
Could someone please advise on this issue?
This question is clearly misrepresenting your actual code/data.
When your database table has 2 rows, but you are receiving 4 rows then the onus is not on the fetching function, but on your query or database table data.
If your actual query is as posted in your question, then your table data contains more than two rows.
If your actual query is different from what you posted (say, joining the table with a copy of itself), then your data is fine and your query is failing you.
Regardless of if you are using mysqli_fetch_array($result), mysqli_fetch_array($result,MYSQLI_ASSOC), or mysqli_fetch_assoc($result), your while() loop will only do one iteration for each row of data.
The difference in resultset fetching functions:
mysqli_fetch_array($result):
array(0=>'1', 'id'=>'1', 1=>'Android', 'deviceType'=>'Android') // 1 row w/ 4 elements
array(0=>'2', 'id'=>'2', 1=>'iPhone', 'deviceType'=>'iPhone') // 1 row w/ 4 elements
mysqli_fetch_array($result,MYSQLI_ASSOC), or mysqli_fetch_assoc($result):
array('id'=>'1', 'deviceType'=>'Android') // 1 row w/ 2 elements
array('id'=>'2', 'deviceType'=>'iPhone') // 1 row w/ 2 elements
I will rewrite your code and implement some good practices:
if($result=mysqli_query($db_conx,"SELECT `deviceType` FROM `table` ORDER BY `id`;")){
if(mysqli_num_rows($result)){
while($row=mysqli_fetch_assoc($result)){
echo ($row["deviceType"]=="Android"?"Android":"IOS"); // inline condition is a personal preference
}
}else{
echo "No rows in `table`.";
}
}
Only bother declaring a variable if you will use its value more than once (*or if it dramatically improves readability to separate it from its single use.)
So that your variable names are intuitive, name your query variable $sql or $query; and name your query's result variable $result.
Only SELECT columns that you intend to use; * is unnecessary for your case.
Backtick ` wrapping is not required on column and table names, but doing so will avoid any potential clashes with MySQL keywords.
Perform a conditional check and declare the $result variable as false or [resultset] in a single step.
Always check that $result is true before calling any functions that access the resultset. (e.g. mysqli_num_rows() and mysqli_fetch_assoc()).
if(mysqli_num_rows($result)){ will check for a non-falsey value -- I mean 0 equates to false and anything greater than 0 will be true.
Your code appears to be perfectly fine. However, instead of the expected output, you get more items than needed. If I am not mistaken, this means that you have duplicate deviceType in your database table. $productCount probably has a value of 4. You can get two values if you use this query instead:
SELECT DISTINCT `deviceType` FROM `table` ORDER BY `id`
but while this should fix the output you get, your data will still hold duplicates, so you might want to look into the data of your table and into the way it was created, find out and fix the problem.
the answer is very simple you are fetching the results twice with the while loop change this line
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
to
while($row = mysqli_fetch_assoc($query)){
then it will work right, you can see buy the order iPhone iPhone android android that is doing it twice instead of once per loop
I've run into a problem that is making me go a bit crazy. I have imported some csv data into a table in my phpadmin database and am now using a php script with mysql_query() to run a simple select query on the database and convert the result into json format - e.g. SELECT clients FROM TABLE 29.
Basically, some of the columns in the table result in a json string after passing them through mysql_query() but others simply return a blank. I have fiddled for hours now and can't figure out why this is. The last bit of my code looks like this:
$myquery = "SELECT `clients` FROM `TABLE 29`";
$query = mysql_query($myquery) or die(mysql_error());
if ( ! $query ) {
echo mysql_error();
die;
}
$data = array();
for ($x = 0; $x < mysql_num_rows($query); $x++) {
$data[] = mysql_fetch_assoc($query);
}
echo json_encode($data);
mysql_close($server);
Any help would be greatly appreciated. Could it be something about the data in the table? I'm at a loss.
thank you!
UPDATE: the length of the strings in the column clients seems to be having an effect. When I replace all the text with something shorter (e.g. aaa instead of something like company name 111 - 045 - project name - currency - etc) it works. However, I need it to be able to handle long strings as I want it to just take whatever users happen to import into it... what am I doing wrong?
No, its not about the table, its about how you loop them. Example:
$data = array();
while($row = mysql_fetch_assoc($query)) { // While a row of data exists, put that row in $row as an associative array
$data[] = $row;
}
echo json_encode($data);
mysql_close($server);
exit;
Note: mysql is depreacted and no longer maintained. Use the improved version of the mysql extension which is mysqli or use PDO instead.
After checking all the rows of the data I discovered that the source of the problem was a 'é' - yes, an 'e' with an accent. Once I replaced it with a regular 'e' the problem went away. So much lost time for something so tiny :(
I am trying to get a count and I am getting 1 instead of 0 from it. I have looked thoroughly though the web and this site. I have even been trying to figure it out on my own for a long time. I keep coming empty handed here.
So Basically what I am trying to do is make a like system for my users. I can get everything to work correctly the count works except for one thing. When they have liked it it returns 1 not 0 which it should be.
Here is my code for the count. I am not posting all the coding for security reasons and it really doesn't need to since its about the counting part not the rest.
$sql_like = mysql_query("SELECT * FROM posts WHERE mem2_id='$id' ORDER BY post_date DESC LIMIT 40");
while($row = mysql_fetch_array($sql_like)){
$like1 = $row['like_array'];
$like3 = explode(",", $like1);
$likeCount = count($like3);
}
So here is the code that determines the number. Any ideas what is wrong with this? Why its returning 1 not 0 when the item is empty?
// you do escape your id right??? (sql injection prevention)
$sql_like = mysql_query("SELECT * FROM posts WHERE mem2_id='$id' ORDER BY post_date DESC LIMIT 40");
while($row = mysql_fetch_array($sql_like)){
$likeCount = 0;
$like1 = trim($row['like_array']);
if ($like1) {
$like3 = explode(",", $like1); // exploding emtpy string would result in array('')
$likeCount = count($like3);
}
}
Calling explode on an empty string gives an array containing the empty string. This is one element, not zero.
I would suggest that you change your database design if possible so that you don't store the values separated by commas. Use a separate table instead.
If you can't change the database design you can handle the empty string separately.
count() returns the number of indexes in an array or something in an object (PHP: count - Manual).
if a string var is used rather than an array or object it returns 1. it has to get a null value in order to return 0.
you can give it a go by trying:
print count("");
and
print count(null);
You'll have better luck if you use explode() to break the sql output into an array and then run a check with an if statement. Something like the following:
$like3 = explode(',',$like1);
if (count($like1)=1 && $like1[0] == '')
// etc ..
I hope this helps