Lets say I have an email. Then I have an input with like 6, on database should be now 6, then I add another value like 11 and now it should be 17. I have like no ideia on how to do this.
The closest thing I did was sum the entire column but that only works for 1.
<?php
$conn = mysqli_connect('localhost','root','','form');
$result = mysqli_query($conn, "SELECT sum(sumemail) FROM data");
$lines = mysqli_num_rows($result);
while($lines = mysqli_fetch_array($result)){
echo "<br><br>Total de fotocópias:";
echo $lines['sum(sumemail)'].'<br/>';
}
?>
I wanted to use this code but i dont know how
UPDATE sumemail = sumemail + $value
I expect to have like 10 emails, all with different values each.
So you just need to know how to use the UPDATE query:
UPDATE <Table> SET sumemail = sumemail + '$value'
Related
hi i have a backend with php in cpanel and i have a problem with one of jsons . this is part of my php code :
...
}elseif ($work == "dollardate") {
$query3 = "SELECT * FROM tabl_dollar_date";
$result3 = $connect->prepare($query3);
$result3->execute();
$out3 = array();
while ($row3 = $result3->fetch(PDO::FETCH_ASSOC)) {
$record3 = array();
$record3["dollar"] = $row3["dollar"];
$record3["date"] = $row3["date"];
array_push($out3, $record3);
}
echo json_encode($out3);
}
?>
this code show this in json :
[
{
"dollar":"15000",
"date":"1397-12-12"
}
]
how can remove array from json and show the json like this :
{
"dollar":"15000",
"date":"1397-12-12"
}
Easiest way (according his code):
change line
echo json_encode($out3);
to
echo json_encode($out3[0]);
One solution is that if you just want the latest value (in case there are multiple records in the table), then change the SELECT to order by date descending also set LIMIT to 1 to only get the 1 record anyway, and remove the loop to fetch the data and just fetch the 1 record...
$query3 = "SELECT `date`, `dollar`
FROM `tabl_dollar_date`
ORDER BY `date` desc
LIMIT 1";
$result3 = $connect->prepare($query3);
$result3->execute();
$row3 = $result3->fetch(PDO::FETCH_ASSOC);
echo json_encode($row3);
As you know which fields you want from the SELECT, it's good to just fetch those fields rather than always using *. This also means that as the result set only contains the fields your after, you can directly json_encode() the result set rather than extracting the fields from one array to another.
I have created a catalog with a block of years search function for characters and the years I've assigned to them. So anything from 1940-1949 would be in the 1940's block of time, and so on. I'm using a href to group these timeframes.
<?php
$sql = "SELECT * FROM catalog";
$displayby = $_GET['displayby'];
$displayvalue = $_GET['displayvalue'];
if($displayby && $displayvalue){
$sql = "SELECT * FROM catalog WHERE $displayby LIKE '$displayvalue'";
}
if($displayby == 'year'){
$min = $_GET['min'];
$max = $_GET['max'];
$sql = "SELECT * FROM catalog WHERE year BETWEEN '$min' AND '$max'";
}
//$result = mysqli_query($con,$sql);
$result = mysqli_query($con,"SELECT * FROM catalog WHERE year BETWEEN '$min' AND '$max'");
while($row = mysqli_fetch_array($result)){
$name = $row['name'];
$filename = $row['filename'];
$cid = $row['cid'];
echo "\n<div class=\"holder\">";
echo "<img src=\"thumbs/$filename\">";
echo "$name<br />\n";
echo "</div>";
}
?>
With this href to only bring up certain characters within those years:
40's Villans<br/>
However they are showing up in the years prior - it might be 1945 as a set date for the character but they only appear in 1930's link.
What am i doing wrong?
Edit : here is the table
BETWEEN ... AND .... can be used for integers and strings. so:
BETWEEN 1 AND 3
Will be true for 1, 2 and 3. But:
BETWEEN 'a' AND 'c'
Will be true for 'a', 'b' and 'c'.
http://dev.mysql.com/doc/refman/5.0/en/comparison-operators.html#operator_between
You've put quotes around your numbers, turning them into strings. So the comparison is done on the characters, not the numbers.
Removing the quotes should make it work.
$result = mysqli_query($con,"SELECT * FROM catalog WHERE year BETWEEN $min AND $max");
But as said by others, there are a lot of other problems with the code. This mistake is a symptom of someone who doesn't really know what they're doing. Why not start at the beginning, and grab a good book. Read it. Do the examples. Experiment. It can take years to become a fluent programmer. (This is meant as an ecouragement, not critizm.)
instead of using between you can also use the following statement for getting the desired data as following
$sql = "SELECT * FROM catalog WHERE year >= '$min' AND year <='$max'";
and in phpmyadmin you can run the query to make sure that you are getting the correct data ,and after that you can check your code where you are displaying it
i hope it help's you..
In my mysql tables I have for example a row that holds values, lets call this table "days" and the row "haircuts".
So I have 5 rows all with "1","2","3","4","5" under "haircuts". What I want to do, in php, is add them all together and echo. So it would maybe look like this:
echo "Total haircuts: ",$haircuts;
Which would hopefully show:
Total haircuts: 15
I can't figure out how, though. Here's how I'm currently doing it.
$getstats = mysqli_query($con,"SELECT * FROM stats");
$gotstats = mysqli_fetch_array($getstats);
$haircuts = $gotstats['haircuts'];
But it only echoes the "haircuts" of the first row? :(
I have multiple columns I would like to total and echo too if that is possible?
You can use the MySQL statment SUM in your query, just change your code by this one:
$getstats = mysqli_query($con,"SELECT SUM(haircuts) AS totalhaircuts FROM stats");
$gotstats = mysqli_fetch_array($getstats);
$totalhaircuts = $gotstats['totalhaircuts'];
echo "Total haircuts: ",$totalhaircuts;
You probably should use MySQL SUM().
SELECT SUM(haircuts) as tot_cuts FROM stats;
$gotstats['tot_cuts'];
You need to use loop for getting all rows. Example:
$getstats = mysqli_query($con,"SELECT * FROM stats");
while($gotstats = mysqli_fetch_array($getstats)){
$haircuts = $gotstats['haircuts'];
echo $haircuts . '<br />';
}
I have to identify unsaved value from mysql table using php and mysql, for example i using table named as numtab and i have already stored some numbers 1, 3, 4, 7, 23, 12, 45 in numb column.
now i have generated one new number randomly(for example 23) and i have to check this number with already stored numbers,
if 23 is exist in the table mean i have to generate another one new number and have to check once again with stored values, this process have to continue till finding unsaved number.
if generated value is not exist in table mean can stop the process and can store this number in table.
here below the format i am currently using
$numb=23;
$qryb="select * from numtab where numb='$numb'";
$results=mysql_query($qryb)or die("ERROR!!");
if(mysql_num_rows($results) == 1)
{
$numb=rand(1,100);
mysql_query("insert query");
}
the problem is above the code is validation once only, its not verifying second time. i think if using for or while loop mean can solve this problem, but i dont know how to do looping, so help me to solve this problem.
You can use in clause like this :
$qryb="select * from numtab where numb in('$numb')";
$results=mysql_query($qryb)or die("ERROR!!");
$count = mysql_num_rows($results);
if ($count > 0) {
echo "number exist in db";
} else {
echo "number does not exist in db";
}
You could make a while() loop to check if the numbers exist in your database. You could also retrieve all numbers from the database, store them in an array and check if the generated number exists within that array.
The first option would be something like this:
do {
$numb=rand(1,100);
$qryb="select * from numtab where numb='$numb'";
$results = mysql_query($qryb) or die("ERROR!!");
} while(mysql_num_rows($results) >= 1)
mysql_query("insert query");
The second option would be something like this:
$query = mysql_query("SELECT DISTINCT(numb) as numb FROM numtab");
// set array
$array = array();
// look through query
while($row = mysql_fetch_assoc($query)){
// add each row returned into an array
$array[] = $row['numb'];
}
do
{
$numb = rand(1,100);
}
while(!in_array ( $numb , $array) ;
mysql_query("insert query");
$query = "select Code , count(ListID) as nums from accesstable where Cust=" . $_SESSION ['Cust'] . " and App=" . $_SESSION ['App'] . " group by Code";
$result = mysql_query ( $query );
while ($row = mysql_fetch_array ( $result )){
$Codes[] = $row['Code'];
$Values[] = $row['nums'];
}
This is the structure of my code that I am trying to learn how to properly access... Here is my dilemma... I am trying to figure out how to explicitly find the associated count of nums dependent on the value of a Code.
Let me explain in better detail where my issue is....
Lets say the list of codes is
Code nums
1 624
7 825
571 450
9 393
2 739
9 590
The above code does successfully allow me to separate those values strictly into keys and values but I cannot figure out how to grab the nums value if the code is = to a certain value... I have currently been trying to declare a variable above the entire snippet of code and then declare it within the while statement but cannot figure out how to get the value to bind properly.... I will repaste the above code with one of my many failures in the while statement to give a better idea.
$Answer1 = 0;
$query = "select Code , count(ListID) as nums from accesstable where Cust=" . $_SESSION ['Cust'] . " and App=" . $_SESSION ['App'] . " group by Code";
$result = mysql_query ( $query );
while ($row = mysql_fetch_array ( $result )){
$Codes[] = $row['Code'];
$Values[] = $row['nums'];
($Codes == 1){
$Answer1 = // Right Here I want to Get the value 624 related to Code 1... Dont want to embarass myself with examples of what I have tried...
}
So how do I make a condition to output the value associated with a Code? I want to explicitly define these values as the list of codes can change with each customer... Luckily there are only a certain amount of codes so its not like I need to define too many of them... I just want to make sure I can get the nums value associated with a code and display it.
Hope I did a good job explaining this. :)
I'd do:
while ($row = mysql_fetch_array ( $result )){
$Codes[] = $row['Code'];
$Values[$row['Code']] = $row['nums'];
}
and, to access the value associated to a code:
$code = 1;
$value = $values[$code];
Since they would share the same array key, something like this would work-
if ($Codes[$key] == 1){
$Answer1 = $Values[$key];
}