PHP mysql session array
I have a little question.
I have few stored values in session Array. These values are ID's of products.
After that i want to display products from my database, but this is not working propertly for me.
Can someone help me a little? :) (I am still learning :) )
<?php
include 'includes/dbconnect.php';
$orderid = $_SESSION['order'];
foreach ($orderid as $value) {
$sql="SELECT * FROM product WHERE productID LIKE '%$value%'";
$result=$conn->query($sql);
while($row=$result->fetch_assoc()){
echo '<tr>';
echo '<td>'.$row["tag"].'</td>';
echo '<td>'.$row["price"].',- Kč</td>';
echo '<td><a href="product.php?id='.$row["productID"].'"><img src="images/'.$row["tag"].'.jpg" width=70"></a></td>';
echo '<td>1</td>' ;
echo '<td><a href="#" class="btn btn-danger btn-lg">X</a></td>';
}
}
?>
var_dump($orderid); shows:
array(1) {
["order"]=> array(10) {
[0]=> string(2) "44"
[1]=> string(2) "46"
[2]=> string(2) "44"
[3]=> string(2) "54"
[4]=> string(1) "1"
[5]=> string(2) "44"
[6]=> string(1) "1"
[7]=> string(2) "44"
[8]=> string(2) "47"
[9]=> string(2) "74"
}
}
php mysql session
edited Nov 21 '18 at 16:45
Fanie Void
1619
asked Nov 21 '18 at 15:55
Pe TrPe Tr
62
|
show 9 more comments
I have a little question.
I have few stored values in session Array. These values are ID's of products.
After that i want to display products from my database, but this is not working propertly for me.
Can someone help me a little? :) (I am still learning :) )
<?php
include 'includes/dbconnect.php';
$orderid = $_SESSION['order'];
foreach ($orderid as $value) {
$sql="SELECT * FROM product WHERE productID LIKE '%$value%'";
$result=$conn->query($sql);
while($row=$result->fetch_assoc()){
echo '<tr>';
echo '<td>'.$row["tag"].'</td>';
echo '<td>'.$row["price"].',- Kč</td>';
echo '<td><a href="product.php?id='.$row["productID"].'"><img src="images/'.$row["tag"].'.jpg" width=70"></a></td>';
echo '<td>1</td>' ;
echo '<td><a href="#" class="btn btn-danger btn-lg">X</a></td>';
}
}
?>
var_dump($orderid); shows:
array(1) {
["order"]=> array(10) {
[0]=> string(2) "44"
[1]=> string(2) "46"
[2]=> string(2) "44"
[3]=> string(2) "54"
[4]=> string(1) "1"
[5]=> string(2) "44"
[6]=> string(1) "1"
[7]=> string(2) "44"
[8]=> string(2) "47"
[9]=> string(2) "74"
}
}
php mysql session
edited Nov 21 '18 at 16:45
Fanie Void
1619
asked Nov 21 '18 at 15:55
Pe TrPe Tr
62
1
add the error message(s) you get.We cannot debug this for you without more information on what went wrong
– Akintunde-Rotimi
Nov 21 '18 at 15:58
3
i don't see session_start()
– suresh bambhaniya
Nov 21 '18 at 15:59
1
In $sql query instead of usingLIKE '%$value%'";try this:LIKE '". $value ."'";
– Studocwho
Nov 21 '18 at 16:10
1
OHHH thanks Studocwho :) thats it!! :) thanks alot :)
– Pe Tr
Nov 21 '18 at 16:12
1
why searching complicated when it's simple ? x) @Studocwho then you just have to make it an answer
– Fanie Void
Nov 21 '18 at 16:16
|
show 9 more comments
I have a little question.
I have few stored values in session Array. These values are ID's of products.
After that i want to display products from my database, but this is not working propertly for me.
Can someone help me a little? :) (I am still learning :) )
<?php
include 'includes/dbconnect.php';
$orderid = $_SESSION['order'];
foreach ($orderid as $value) {
$sql="SELECT * FROM product WHERE productID LIKE '%$value%'";
$result=$conn->query($sql);
while($row=$result->fetch_assoc()){
echo '<tr>';
echo '<td>'.$row["tag"].'</td>';
echo '<td>'.$row["price"].',- Kč</td>';
echo '<td><a href="product.php?id='.$row["productID"].'"><img src="images/'.$row["tag"].'.jpg" width=70"></a></td>';
echo '<td>1</td>' ;
echo '<td><a href="#" class="btn btn-danger btn-lg">X</a></td>';
}
}
?>
var_dump($orderid); shows:
array(1) {
["order"]=> array(10) {
[0]=> string(2) "44"
[1]=> string(2) "46"
[2]=> string(2) "44"
[3]=> string(2) "54"
[4]=> string(1) "1"
[5]=> string(2) "44"
[6]=> string(1) "1"
[7]=> string(2) "44"
[8]=> string(2) "47"
[9]=> string(2) "74"
}
}
php mysql session
edited Nov 21 '18 at 16:45
Fanie Void
1619
asked Nov 21 '18 at 15:55
Pe TrPe Tr
62
I have a little question.
I have few stored values in session Array. These values are ID's of products.
After that i want to display products from my database, but this is not working propertly for me.
Can someone help me a little? :) (I am still learning :) )
<?php
include 'includes/dbconnect.php';
$orderid = $_SESSION['order'];
foreach ($orderid as $value) {
$sql="SELECT * FROM product WHERE productID LIKE '%$value%'";
$result=$conn->query($sql);
while($row=$result->fetch_assoc()){
echo '<tr>';
echo '<td>'.$row["tag"].'</td>';
echo '<td>'.$row["price"].',- Kč</td>';
echo '<td><a href="product.php?id='.$row["productID"].'"><img src="images/'.$row["tag"].'.jpg" width=70"></a></td>';
echo '<td>1</td>' ;
echo '<td><a href="#" class="btn btn-danger btn-lg">X</a></td>';
}
}
?>
var_dump($orderid); shows:
array(1) {
["order"]=> array(10) {
[0]=> string(2) "44"
[1]=> string(2) "46"
[2]=> string(2) "44"
[3]=> string(2) "54"
[4]=> string(1) "1"
[5]=> string(2) "44"
[6]=> string(1) "1"
[7]=> string(2) "44"
[8]=> string(2) "47"
[9]=> string(2) "74"
}
}
php mysql session
php mysql session
edited Nov 21 '18 at 16:45
Fanie Void
1619
asked Nov 21 '18 at 15:55
Pe TrPe Tr
62
edited Nov 21 '18 at 16:45
Fanie Void
1619
asked Nov 21 '18 at 15:55
Pe TrPe Tr
62
edited Nov 21 '18 at 16:45
Fanie Void
1619
edited Nov 21 '18 at 16:45
Fanie Void
1619
edited Nov 21 '18 at 16:45
Fanie Void
1619
1619
asked Nov 21 '18 at 15:55
Pe TrPe Tr
62
asked Nov 21 '18 at 15:55
Pe TrPe Tr
62
asked Nov 21 '18 at 15:55
Pe TrPe Tr
62
62
1
add the error message(s) you get.We cannot debug this for you without more information on what went wrong
– Akintunde-Rotimi
Nov 21 '18 at 15:58
3
i don't see session_start()
– suresh bambhaniya
Nov 21 '18 at 15:59
1
In $sql query instead of usingLIKE '%$value%'";try this:LIKE '". $value ."'";
– Studocwho
Nov 21 '18 at 16:10
1
OHHH thanks Studocwho :) thats it!! :) thanks alot :)
– Pe Tr
Nov 21 '18 at 16:12
1
why searching complicated when it's simple ? x) @Studocwho then you just have to make it an answer
– Fanie Void
Nov 21 '18 at 16:16
|
show 9 more comments
1
add the error message(s) you get.We cannot debug this for you without more information on what went wrong
– Akintunde-Rotimi
Nov 21 '18 at 15:58
3
i don't see session_start()
– suresh bambhaniya
Nov 21 '18 at 15:59
1
In $sql query instead of usingLIKE '%$value%'";try this:LIKE '". $value ."'";
– Studocwho
Nov 21 '18 at 16:10
1
OHHH thanks Studocwho :) thats it!! :) thanks alot :)
– Pe Tr
Nov 21 '18 at 16:12
1
why searching complicated when it's simple ? x) @Studocwho then you just have to make it an answer
– Fanie Void
Nov 21 '18 at 16:16
1
1
add the error message(s) you get.We cannot debug this for you without more information on what went wrong
– Akintunde-Rotimi
Nov 21 '18 at 15:58
add the error message(s) you get.We cannot debug this for you without more information on what went wrong
– Akintunde-Rotimi
Nov 21 '18 at 15:58
3
3
i don't see session_start()
– suresh bambhaniya
Nov 21 '18 at 15:59
i don't see session_start()
– suresh bambhaniya
Nov 21 '18 at 15:59
1
1
In $sql query instead of using
LIKE '%$value%'"; try this: LIKE '". $value ."'";– Studocwho
Nov 21 '18 at 16:10
In $sql query instead of using
LIKE '%$value%'"; try this: LIKE '". $value ."'";– Studocwho
Nov 21 '18 at 16:10
1
1
OHHH thanks Studocwho :) thats it!! :) thanks alot :)
– Pe Tr
Nov 21 '18 at 16:12
OHHH thanks Studocwho :) thats it!! :) thanks alot :)
– Pe Tr
Nov 21 '18 at 16:12
1
1
why searching complicated when it's simple ? x) @Studocwho then you just have to make it an answer
– Fanie Void
Nov 21 '18 at 16:16
why searching complicated when it's simple ? x) @Studocwho then you just have to make it an answer
– Fanie Void
Nov 21 '18 at 16:16
|
show 9 more comments
1 Answer
1
active
oldest
votes
Just for the purposes of SO, I'll make my comment as an answer:
In the $sql query instead of using LIKE '%$value%'"; use this: LIKE '". $value ."'";
This ensures that we actually get the value of the variable.
answered Nov 21 '18 at 16:21
StudocwhoStudocwho
1,04811120
add a comment |
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Just for the purposes of SO, I'll make my comment as an answer:
In the $sql query instead of using LIKE '%$value%'"; use this: LIKE '". $value ."'";
This ensures that we actually get the value of the variable.
answered Nov 21 '18 at 16:21
StudocwhoStudocwho
1,04811120
add a comment |
Just for the purposes of SO, I'll make my comment as an answer:
In the $sql query instead of using LIKE '%$value%'"; use this: LIKE '". $value ."'";
This ensures that we actually get the value of the variable.
answered Nov 21 '18 at 16:21
StudocwhoStudocwho
1,04811120
add a comment |
Just for the purposes of SO, I'll make my comment as an answer:
In the $sql query instead of using LIKE '%$value%'"; use this: LIKE '". $value ."'";
This ensures that we actually get the value of the variable.
answered Nov 21 '18 at 16:21
StudocwhoStudocwho
1,04811120
Just for the purposes of SO, I'll make my comment as an answer:
In the $sql query instead of using LIKE '%$value%'"; use this: LIKE '". $value ."'";
This ensures that we actually get the value of the variable.
answered Nov 21 '18 at 16:21
StudocwhoStudocwho
1,04811120
answered Nov 21 '18 at 16:21
StudocwhoStudocwho
1,04811120
answered Nov 21 '18 at 16:21
StudocwhoStudocwho
1,04811120
answered Nov 21 '18 at 16:21
StudocwhoStudocwho
1,04811120
1,04811120
add a comment |
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1
add the error message(s) you get.We cannot debug this for you without more information on what went wrong
– Akintunde-Rotimi
Nov 21 '18 at 15:58
3
i don't see session_start()
– suresh bambhaniya
Nov 21 '18 at 15:59
1
In $sql query instead of using
LIKE '%$value%'";try this:LIKE '". $value ."'";– Studocwho
Nov 21 '18 at 16:10
1
OHHH thanks Studocwho :) thats it!! :) thanks alot :)
– Pe Tr
Nov 21 '18 at 16:12
1
why searching complicated when it's simple ? x) @Studocwho then you just have to make it an answer
– Fanie Void
Nov 21 '18 at 16:16