I have this code
require("connession.php");
$contabilita="d";
$dataa="10/12/2018";
$datam_0=explode("/", $dataa);
$datam = "".$datam_0[2];
$result = mysqli_query($con,"SELECT Max(n_ricevuta)+1 as max_ricevuta
FROM corrispettivi_mod
where anno='$datam'
and contabilita='$contabilita'");
$row = mysqli_fetch_assoc($result);
if(mysqli_num_rows($result) > 0){
echo $row['max_ricevuta'];
}
else{echo "1";}
here is my db
data anno n_ricevuta contabilita
2019-12-01 2019 1 nd
in my db I have only a row with a value of the filed 'anno' 2019 so it must print '1' but it doesn't.
if I modify in the db the field anno with the value "2018", it works.
help me thanks
When using an aggregate function like max() you will always get a row, if there isn't a matching row it will return null.
So you need to change your test to...
if($row['max_ricevuta'] != null){
echo $row['max_ricevuta'];
}
else {
echo "1";
}
As RiggsFolly also points out, worth checking the values your using in your testing, you have
$contabilita="d";
whereas the example data shows nd.
Related
This code only reads the first value present in the column. If the value posted in the html form matches the first value, it inserts into the database. But I want to check all the values in the column and then take the respective actions.
For example, if i give input for 'ppincode' and 'dpincode' as 400001, it accepts. but if i gave 400002, 400003,..... it displays the alert even if those value are present in the database
DATABASE:
pincode <== column_name
400001 <== value
400002
400003
400004
...
also i tried this
$query = "SELECT * FROM pincodes";
$result = mysqli_query($db, $query);
$pincodearray = array();
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)){
$pincodearray[] = $row;
}
}
If I understand well - you want to compare value from POST request with all retrieved records saved in DB and if it matches - perform action.
If so, I would recommend using for(each) loop. Example:
if( !empty($row){
foreach( $row as $key ){
if($key['pincode'] == $ppincode && $key['pincode'] == $dpincode){
// your action goes here
}
}
}
Additional tip: use prepared statements :)
SELECT count(*) FROM table WHERE ppincode=ppincode AND bpincode=bpincode
if this return 0 then insert or else show alert.
hi i have a backend with php in cpanel and i have a problem with one of jsons . this is part of my php code :
...
}elseif ($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.
Have the following php query:
if ($result = mysqli_query($conn, "SELECT scholarship1, scholarship2, scholarship3, scholarship4, scholarship5, scholarship6, scholarship7, scholarship8, scholarship9, scholarship10, date1, date2, date3, date4, date4, date5, date6, date7, date8, date9, date10 FROM scholarships")) {
$row_cnt = mysqli_num_rows($result);
if (($row_cnt) > 0) {
while(mysqli_fetch_assoc($result)) {
if ((['scholarship4'] == null) AND (['scholarship3'] != null)) {
echo "YES";
}
else {
echo "NO";
}
}
}
}
Connection to the database was successful.
$row_cnt > 0 recognizing data in database.
My if statement returns "NO" when 'scholarship4' is clearly NULL and 'scholarship3' clearly has data.
I essentially copied this code from another of my pages that was working properly, so I am flabbergasted as to what is going on here.
One thing I did remove was while($row = ...) since the original data input was not of the 'select' variety. Also, please keep in mind, this is just testing the if statement initially, whereas many if statements are currently hidden from the loop.
Please let me know if more information is needed. Thanks for any and all help.
When you fetch the data, you need to store it and then reference this array in your test...
while($row = mysqli_fetch_assoc($result)) {
if (($row['scholarship4'] == null) AND ($row['scholarship3'] != null)) {
I am having some trouble with displaying some SQL query results.
Goal: I want to display the Helper 'name' in the table that is being generated if there is a helper signed up in the 'signup' table for that event 'eid' (event id).. If (1)there is no helper then display 'waiting for help', (2) there is a helper then display 'name -- awaiting approval..' and (3) else just display the name of helper..
Tried running the SQL query in phpMyAdmin with hard coded values and I get the results that I want so I know it is not my query. Have a suspicion that it is just the print out of the info into the table that is wrong somewhere. The table will display the data up until the ZIP from the address and then the next column which is the 'Helper' column does not display anything at all. So it makes me think I have a simple typo somewhere based on my if() statement logic BUT also find it interesting also that when I do the line:
echo "testing method -> ".getHelperIdOrName(2, 80)."<br>";
I cant get the table to print out at all. Not sure if this is related to my exact issue but it seems it could be. After I put this function in stuff stopped working so it seems like it could be culprit. The return of the function should either return an ID (int), a name "string", or just a generic value X (string)..
Any and all help is appreciated!
function getHelperIdOrName($x, $eid){
//Get the helper name first
$helperName = "";
$helperId = 0;
$sql = "SELECT id, first FROM users WHERE id IN (SELECT helper FROM signup WHERE gner = '".$userId."' AND eid = '".$eid."')";
$result = mysqli_query($db,$sql);
$row = $result->fetch_assoc();
if ($x == 2){
$helperName = $row["first"];
return $helperName;
}
else if ($x == 1){
$helperId = $row["id"];
return $helperId;
}
else {
return "X";
}
}
echo "testing method -> ".getHelperIdOrName(2, 80)."<br>";
//look for calendar and/or business approved events (approved=1) to display on page
$sql = "SELECT s.gner, s.helper, s.eid, s.approved, e.name, e.date, e.summary, e.street, e.city, e.state, e.zip
FROM signup s
INNER JOIN events e ON e.id = s.eid
INNER JOIN users u ON u.id = s.gner
WHERE s.gner = '".$userId."'";
$result = mysqli_query($db,$sql);
echo "<h3 class=\"text-center\">Events I'm Going To</h3>";
echo "<table class=\"table table-hover\"><tr><th>Event Name</th><th>Date</th><th>Summary</th><th>Location</th><th>Helper</th><th>Remove</th></tr>";
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>".$row["name"]."</td><td>".$row["date"]."</td><td>".$row["summary"]."</td><td>".$row["street"].", "
.$row["city"].", ".$row["state"]." ".$row["zip"]."</td>";
$tmp_eid = $row["eid"];
if (getHelperIdOrName(2, $temp_eid) == "X"){
echo "<td>Waiting for help..</td>";
}
else if ($row["approved"] == 0){
echo "<td>".getHelperIdOrName(2, $temp_eid)." -- Awaiting Approval (see below)</td>";
}
else {
echo "<td>".getHelperIdOrName(2, $temp_eid)."</td>";
}
echo "<td><form method=\"post\" action=\"remove.php\">
<button type=\"submit\" name=\"remove\" value=\"".$row["eid"]."\">Not Going</button></form></td></table>";
}
}
else echo "</table><br><p class=\"text-center\">You are not signed up for any events. Click here to sign up for events near you!</p>";
Thanks for that Jeff. The issue was that inside of the function it indeed did not know what $userId was even though I had the include statement at the top of my php file. I had to add this line into my function at the top..
global $db; //is part of my db my connection info in my config.php file
and then I also passed the $userId to the function as a parameter
these lines are what I used to help me see the errors:
ini_set('display_startup_errors', 1);
ini_set('display_errors', 1);
error_reporting(-1);
i also had some ending < /table > tags inside some if logic so that fixed the funky displays I was getting (2nd row of table being outside of the table)
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");