is there an issue with PHP not allowing multiple function parameters and returning the correct value. Here is the code:
function getConfig($name) {
$sql = "SELECT value FROM config WHERE name = '".$name."'";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
print $row["value"];
}
getConfig("name");
While the code above works, the code bellow similar does not work. Here is the code:
function getConfig($name, $from) {
$sql = "SELECT value '".$from."' config WHERE name = '".$name."'";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
print $row["value"];
}
getConfig("name", "config");
Why does that occur that the 1st function works and the second does not?
$sql = "SELECT value FROM '".$from."' WHERE name = '".$name."'";
// FROM TABLE_NAME
// and not TABLE_NAME TABLE_NAME
Related
I have written a code in PHP and not able to get a response from the code. Following is my code.
<?php
include_once("conectionFile.php");
$db = new Data();
$con = $db->Conn();
$val = $_REQUEST;
$lclQuery = "SELECT * FROM student_details WHERE st_id = ".$val."";
$lclResult = $con->query($lclQuery);
if($row = $lclResult->fetchAll(PDO::FETCH_ASSOC)) {
echo $res = json_encode($row);
}
?>
$lclQuery = "SELECT * FROM student_details WHERE st_id = ".$val."";
You need to change your query as it is not correct also $val is array so you need to use correct key there in order to get value for example $val = $_REQUEST['some_id'] then change query like
$lclQuery = "SELECT * FROM student_details WHERE st_id = '".$val."'";
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);
How can I get a single column value from mysqli? The result should be single row with only one column.
This is what I have tried:
$query = "SELECT MAX(`userid`) FROM `user`";
$rlt = mysqli_query($this->db, $query);
echo $rlt['userid'];
You are not fetching the row after executing the query:
$query = "SELECT MAX(`userid`) FROM `user'";
$rlt = mysqli_query($this->db,$query);
$row = mysqli_fetch_row($this->db, $rlt);
echo $row[0];
The alternative would be to use an alias for the computed field and use fetch_assoc:
$query = "SELECT MAX(`userid`) as `maxid` FROM `user'";
$rlt = mysqli_query($this->db,$query);
$row = mysqli_fetch_assoc($this->db, $rlt);
echo $row['maxid'];
try with create alias and fetch result after query execution also your quote of user' looking wrong
$query = "SELECT MAX(`userid`) as userid FROM `user`";
$rlt = mysqli_query($this->db,$query);
$r = mysqli_fetch_assoc($rlt);
echo $r['userid'];
or fetch only one row like:-
$r = mysqli_fetch_row($this->db, $rlt);
echo $r[0];
You should create alias here.
Use this one:
$query = "SELECT MAX(`userid`) as Max_Userid FROM `user'";
$rlt = mysqli_query($this->db,$query);
$row = mysqli_fetch_assoc($this->db, $rlt);
echo $row['Max_Userid'];
Note: Always use alias when you use mysql function in query with fields.
I got a table named "Serials" with 5 comumns
Serial, Code, Name, Redeemed, Redeem_date
i am selecting some fields from that table with this query:
$query = "SELECT `Name`,`Redeemed`,`Redeem_date` FROM `Serials` WHERE `Serial` = '$serial' AND `Code` = '$code'";
$db->setQuery($query);
$db->query();
But i dont know how to pass these values in the following variables so i can use them in if statements later
$name= //retured value from column Name
$redeemed= //retured value from column Redeemed
$redeem_date= //retured value from column Redeem_date
just like this..
// Your query here..
$query = "SELECT `Name`,`Redeemed`,`Redeem_date` FROM `Serials` WHERE `Serial` = '$serial' AND `Code` = '$code'";
$db->setQuery($query);
$results = $db->query();
//fetch data and stored into variables
while($row = fetch_array($results)){
$name = $row['Name'];
$redeemed = $row['Redeemed'];
$redeem_date = $row['Redeem_date'];
}
try something like this :
<?php
$result = $db->query("SELECT `Name`,`Redeemed`,`Redeem_date` FROM `Serials` WHERE `Serial` = '$serial' AND `Code` = '$code'");
while (list($name, $redeemed, $redeem_date) = $result->fetch(PDO::FETCH_NUM)) {
// DO SOMETHING
}
?>
while ($row = $db->fetch()) {
$name= $row['name'];
$redeemed= $row['redeemed'];
$redeem_date= $row['redeem_date'];
}
this one might fetch your results and assign to vars
I have a query that gets 5 lines of data like this example below
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
}
I want to run a query inside each results like this below
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}
but for some reason it's only display the first line when I add the query inside it. I can seem to make it run all 5 queries.
SELECT
t1.ref,
t1.user,
t1.id,
t2.domain,
t2.title
FROM
table AS t1
LEFT JOIN anothertable AS t2 ON
t2.domain = t1.ref
LIMIT
0, 5
The problem is that inside the while-cycle you use the same variable $result, which then gets overridden. Use another variable name for the $result in the while cycle.
You change the value of your $query in your while loop.
Change the variable name to something different.
Ex:
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$qry = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$rslt = mysql_query($qry) or die(mysql_error());
if (mysql_num_rows($rslt) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}
Use the following :
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query_domain = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result_domain = mysql_query($query_domain) or die(mysql_error());
if (mysql_num_rows($result_domain) )
{
$row_domain = mysql_fetch_row($result_domain);
$title = $row_domain['title'];
} else {
$title = "No Title";
}
echo "$ref - $title";
}
This is a logical problem. It happens that way, because you are same variable names outside and inside the loop.
Explanation:
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
// Now $results hold the result of the first query
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
//Using same $query does not affect that much
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
//But here you are overriding the previous result set of first query with a new result set
$result = mysql_query($query) or die(mysql_error());
//^ Due to this, next time the loop continues, the $result on whose basis it would loop will already be modified
//..............
Solution 1:
Avoid using same variable names for inner result set
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$sub_result = mysql_query($query) or die(mysql_error());
// ^ Change this variable so that it does not overrides previous result set
Solution 2:
Avoid the double query situation. Use joins to get the data in one query call. (Note: You should always try to optimize your query so that you will minimize the number of your queries on the server.)
SELECT
ref,user,id
FROM
table t
INNER JOIN
anothertable t2 on t.ref t2.domain
LIMIT 0, 5
Learn about SQL joins:
SELECT table.ref, table.user, table.id, anothertable.title
FROM table LEFT JOIN anothertable ON anothertable.domain = table.ref
LIMIT 5
You're changing the value of $result in your loop. Change your second query to use a different variable.
it is not give proper result because you have used same name twice, use different name like this edit.
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query1 = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result1 = mysql_query($query1) or die(mysql_error());
if (mysql_num_rows($result1) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}