My code looks like this:
$sql = "SELECT `sw1`, `sw2`, `sw3`, `sw4`, `fb1`, `fb2`, `fb3`, `fb4`, `bew1`, `bew2`, `bew3`, `bew4` FROM `reg` WHERE `id` = ".$id." ORDER BY `id` ASC LIMIT 0, 30 ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
foreach($row as $x => $x_value) {
echo "Key=" . $x . ", Value=" . $x_value;
echo "<br>";
}
}
}
There is some data in sw2 but it isn't shown.
When I tryed to UPDATE data the data in the db wasn't changed.
$id is right.
other data from the table could be read.
This code works fine:
$sql = "SELECT * FROM `reg` WHERE `id` = ".$id." ORDER BY `id` ASC LIMIT 0, 30 ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$row = $result->fetch_assoc();
print_r($row);
}
Try your query with echo.
same trouble follows try changing Where 'id' = "$id"
to Where 'id' like "$id"
You dont need quotes:
$sql = "SELECT sw1, sw2, sw3, sw4, fb1, fb2, fb3, fb4,
bew1, bew2, bew3, bew4
FROM reg WHERE id = $id";
Related
Is it possible to extract highest id from table (in this case 9) and return it as variable $maximum, which I can use later as integer?
You can use MAX() function. Doc can be found here
Try this, It will must work.
$sql = "SELECT id FROM some_table ORDER BY id DESC LIMIT 1";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "id = " . $row["id"];
}
}
You can try with following snippet:
$sql = "SELECT * FROM some_table where id = (select max(id) from some_table)";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "id = " . $row["id"];
}
}
Try this, It will must work.
$sql = "SELECT id FROM some_table ORDER BY id DESC LIMIT 1";
$result = $conn->query($sql);
$id = 0;
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$id = $row["id"];
}
}
echo $id;
I want to display hospital names from database by two ways:
By Selecting city.
By entering the name of hospital in search bar.
I wrote below php script. The 1st part works fine and shows all hospitals from a selected city, but 2nd part doesn't work for me. There is no error displayed. Need help with this. Did i not put the 'if' conditions in the right place? Or I missed something else?
if (isset($_POST['search'])) {
if (isset($_POST['search-by-city'])) {
$city_id = $_POST['search-by-city'];
$query = "SELECT * FROM `hospitals` WHERE `City_ID` LIKE '%$city_id%'";
$result = mysqli_query($con,$query);
if (mysqli_num_rows($result) == 0) {
echo '<h2>No recod Found</h2>' ;
}
}
if (isset($_POST['search-by-name'])) {
$hospital_name = $_POST['search-by-name'];
$query = "SELECT * FROM `hospitals` WHERE `Name` LIKE '%$hospital_name%'";
$result = filterTable($query); {
if (mysqli_num_rows($result) == 0) {
echo '<h2>No recod Found</h2>' ;
}
}
while($row = mysqli_fetch_array($result)){
$city_id = $row[3];
$query = "SELECT `Name` FROM `cities` WHERE `ID` LIKE '$city_id'";
$result2 = mysqli_query($con,$query);
$row2 = mysqli_fetch_row($result2);
$city_name = $row2[0];
echo '
<h3>'.$row[1].'</h3>
<h4>'.$city_name.'</h4>
';
}
}
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);
I have a table with a column called 'status'. The defaut value of 'status' is 0. I want to update the value to '1' after using it.
I basically want to check if the status is 0, if it is, do an operation and then change the value to 1.
Here is the code. All works perfectly except that the value of 0 is not changed to 1.
I am a novice so maybe is a very basic mistake :(
<?php
$sql = "SELECT number, status FROM summonerid";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$SummonerID = $row["number"];
$status = $row["status"];
if($status=='0'){
$recentgames=$lol->getRecentGames($SummonerID);
$MatchID1=$recentgames->games[0]->gameId;
$sql = "INSERT INTO matchid (number) SELECT * FROM (SELECT '$MatchID1') AS tmp WHERE NOT EXISTS (SELECT number FROM matchid WHERE number = '$MatchID1') LIMIT 1;";
$sql = "UPDATE summonerid SET status='1' WHERE status='0';"; // THIS IS THE PART THAT DOES NOT WORK WELL
}
}
}
?>
Any help would be highly appreciated
Try this.. you are not executing the sql statements
$sql = "SELECT number, status FROM summonerid";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$SummonerID = $row["number"];
$status = $row["status"];
if($status=='0'){
$recentgames=$lol->getRecentGames($SummonerID);
$MatchID1=$recentgames->games[0]->gameId;
$sql = "INSERT INTO matchid (number) SELECT * FROM (SELECT '$MatchID1') AS tmp WHERE NOT EXISTS (SELECT number FROM matchid WHERE number = '$MatchID1') LIMIT 1;";
$result1 = $conn->query($sql);
$sql = "UPDATE summonerid SET status='1' WHERE status='0';";
$result1 = $conn->query($sql);
}
}
}
?>
I want to get an id from browser and display some pictures from the database.
If there is no "display2.php?productid=" found, then I want to display default image.
How can I do that?
Here is my code;
$sql = "SELECT * FROM productlist where productid=".$_GET['productid'];
$result = $mysqli->query($sql);
while($myRow = $result->fetch_array())
{
if(null !==($_GET['productid']==$myRow["productid"])){
echo "<img src=".$myRow["productid"].">";
}
else {
echo "<img src="SELECT productimage FROM productlist where productid = 1;">";
}
}
Now I will make it easier to explain for you...
Check this out;
//This part works without any problem
$sql = "SELECT * FROM productlista where productid=".$_GET['productid'];
$result = $mysqli->query($restwo);
while($myRow = $resulttwo->fetch_array())
{
if(null !==($_GET['productid']==$myRow["productid"])){
echo "<img src=".$myRow["productimage"].">";
}
//This part below (that should be default) does not work...
if (!$_GET){
echo "hello world"; }
Asaph pointed out SQL injection. You should bind the parameter (google it), or at the minimum do this:
$defaultImage = "SELECT * FROM productlist WHERE imageSrc != '' OR IS NOT NULL ORDER BY productid DESC LIMIT 1";
// run the query, get the result, create a variable with default image...
$defaultImageSrc = ''; // what you get from the query result
$_GET['productid'] = preg_replace('#[^0-9]#', '', $_GET['productid']);
$sql = "SELECT * FROM productlist where productid=".$_GET['productid'];
$result = $mysqli->query($sql);
while($myRow = $result->fetch_array()) {
if(!$myRow['imageSrc']) $myRow['imageSrc'] = $defaultImageSrc;
echo '<img src="'.$path.'">';
}
If you want either $_GET['productid'] or the max(productid) when $_GET['productid'] is not set, you can use a ternary to change your sql query
$productid = ! empty($_GET['productid']) ? " WHERE productid = ".(int)$_GET['productid'] : " ORDER BY productid DESC LIMIT 1";
$sql = "SELECT * FROM productlist".$productid
$result = $mysqli->query($sql);
while($myRow = $result->fetch_array())
{
echo "<img src=".$myRow["productimage"].">";
}
so if isset($_GET['productid']) your query would be
SELECT * FROM productlist WHERE productid = (int)$_GET['productid']
but if not the default would be
SELECT * FROM productlist ORDER BY productid DESC LIMIT 1