Uncaught Error: Call to a member function fetchAll()
-1
Guys i am trying to do a webservice for uni project. Everything seems ok untill i run the code and get this error Uncaught Error: Call to a member function fetchAll() on null . This is the code can anyone please tell me what is wrong with it?
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$loc = $_GET["location"];
$type = $_GET["type"];
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$loc' and type='$type'")
}
else if (isset($_GET["location"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$loc'");
}
else if (isset($_GET["type"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$type'");
}
$rows = $result->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rows);
?>
php mysql pdo
edited Mar 29 '18 at 21:12
tadman
152k18173207
asked Mar 29 '18 at 20:47
Gabriel GanchevGabriel Ganchev
1
add a comment |
-1
Guys i am trying to do a webservice for uni project. Everything seems ok untill i run the code and get this error Uncaught Error: Call to a member function fetchAll() on null . This is the code can anyone please tell me what is wrong with it?
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$loc = $_GET["location"];
$type = $_GET["type"];
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$loc' and type='$type'")
}
else if (isset($_GET["location"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$loc'");
}
else if (isset($_GET["type"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$type'");
}
$rows = $result->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rows);
?>
php mysql pdo
edited Mar 29 '18 at 21:12
tadman
152k18173207
asked Mar 29 '18 at 20:47
Gabriel GanchevGabriel Ganchev
1
2
If none of theifconditions are true, you never set$result.
– Barmar
Mar 29 '18 at 20:48
2
You also are open to SQL injections. You should parameterize the parameters you are passing in.
– chris85
Mar 29 '18 at 20:51
WARNING: When using PDO you should be using prepared statements with placeholder values and supply any user data as separate arguments. In this code you have potentially severe SQL injection bugs. Never use string interpolation or concatenation and instead use prepared statements and never put$_POST,$_GETor any user data directly in your query. Refer to PHP The Right Way for general guidance and advice.
– tadman
Mar 29 '18 at 21:13
add a comment |
-1
-1
-1
Guys i am trying to do a webservice for uni project. Everything seems ok untill i run the code and get this error Uncaught Error: Call to a member function fetchAll() on null . This is the code can anyone please tell me what is wrong with it?
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$loc = $_GET["location"];
$type = $_GET["type"];
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$loc' and type='$type'")
}
else if (isset($_GET["location"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$loc'");
}
else if (isset($_GET["type"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$type'");
}
$rows = $result->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rows);
?>
php mysql pdo
edited Mar 29 '18 at 21:12
tadman
152k18173207
asked Mar 29 '18 at 20:47
Gabriel GanchevGabriel Ganchev
1
Guys i am trying to do a webservice for uni project. Everything seems ok untill i run the code and get this error Uncaught Error: Call to a member function fetchAll() on null . This is the code can anyone please tell me what is wrong with it?
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$loc = $_GET["location"];
$type = $_GET["type"];
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$loc' and type='$type'")
}
else if (isset($_GET["location"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$loc'");
}
else if (isset($_GET["type"]))
{
$result = $conn->query("Select * from resit_accommodation where location='$type'");
}
$rows = $result->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rows);
?>
php mysql pdo
php mysql pdo
edited Mar 29 '18 at 21:12
tadman
152k18173207
asked Mar 29 '18 at 20:47
Gabriel GanchevGabriel Ganchev
1
edited Mar 29 '18 at 21:12
tadman
152k18173207
asked Mar 29 '18 at 20:47
Gabriel GanchevGabriel Ganchev
1
edited Mar 29 '18 at 21:12
tadman
152k18173207
edited Mar 29 '18 at 21:12
tadman
152k18173207
edited Mar 29 '18 at 21:12
tadman
152k18173207
152k18173207
asked Mar 29 '18 at 20:47
Gabriel GanchevGabriel Ganchev
1
asked Mar 29 '18 at 20:47
Gabriel GanchevGabriel Ganchev
1
asked Mar 29 '18 at 20:47
Gabriel GanchevGabriel Ganchev
1
1
2
If none of theifconditions are true, you never set$result.
– Barmar
Mar 29 '18 at 20:48
2
You also are open to SQL injections. You should parameterize the parameters you are passing in.
– chris85
Mar 29 '18 at 20:51
WARNING: When using PDO you should be using prepared statements with placeholder values and supply any user data as separate arguments. In this code you have potentially severe SQL injection bugs. Never use string interpolation or concatenation and instead use prepared statements and never put$_POST,$_GETor any user data directly in your query. Refer to PHP The Right Way for general guidance and advice.
– tadman
Mar 29 '18 at 21:13
add a comment |
2
If none of theifconditions are true, you never set$result.
– Barmar
Mar 29 '18 at 20:48
2
You also are open to SQL injections. You should parameterize the parameters you are passing in.
– chris85
Mar 29 '18 at 20:51
WARNING: When using PDO you should be using prepared statements with placeholder values and supply any user data as separate arguments. In this code you have potentially severe SQL injection bugs. Never use string interpolation or concatenation and instead use prepared statements and never put$_POST,$_GETor any user data directly in your query. Refer to PHP The Right Way for general guidance and advice.
– tadman
Mar 29 '18 at 21:13
2
2
If none of the
if conditions are true, you never set $result.– Barmar
Mar 29 '18 at 20:48
If none of the
if conditions are true, you never set $result.– Barmar
Mar 29 '18 at 20:48
2
2
You also are open to SQL injections. You should parameterize the parameters you are passing in.
– chris85
Mar 29 '18 at 20:51
You also are open to SQL injections. You should parameterize the parameters you are passing in.
– chris85
Mar 29 '18 at 20:51
WARNING: When using PDO you should be using prepared statements with placeholder values and supply any user data as separate arguments. In this code you have potentially severe SQL injection bugs. Never use string interpolation or concatenation and instead use prepared statements and never put
$_POST, $_GET or any user data directly in your query. Refer to PHP The Right Way for general guidance and advice.– tadman
Mar 29 '18 at 21:13
WARNING: When using PDO you should be using prepared statements with placeholder values and supply any user data as separate arguments. In this code you have potentially severe SQL injection bugs. Never use string interpolation or concatenation and instead use prepared statements and never put
$_POST, $_GET or any user data directly in your query. Refer to PHP The Right Way for general guidance and advice.– tadman
Mar 29 '18 at 21:13
add a comment |
1 Answer
1
active
oldest
votes
3
If neither of the $_GET parameters is set, you never set $result to anything, so you'll get an error if you try to use it.
You should also use a prepared statement rather than substituting variables, to prevent SQL injection.
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$stmt = null;
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc and type= :type");
$stmt->execute(['loc' => $_GET['location'], 'type' => $_GET['type']]);
}
elseif (isset($_GET["location"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc");
$stmt->execute(['loc' => $_GET['location']]);
}
elseif (isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :type");
$stmt->execute(['type' => $_GET['type']]);
}
if ($stmt) {
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
} else {
$rows = ;
}
echo json_encode($rows);
?>
answered Mar 29 '18 at 20:54
BarmarBarmar
420k35244344
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
If neither of the $_GET parameters is set, you never set $result to anything, so you'll get an error if you try to use it.
You should also use a prepared statement rather than substituting variables, to prevent SQL injection.
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$stmt = null;
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc and type= :type");
$stmt->execute(['loc' => $_GET['location'], 'type' => $_GET['type']]);
}
elseif (isset($_GET["location"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc");
$stmt->execute(['loc' => $_GET['location']]);
}
elseif (isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :type");
$stmt->execute(['type' => $_GET['type']]);
}
if ($stmt) {
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
} else {
$rows = ;
}
echo json_encode($rows);
?>
answered Mar 29 '18 at 20:54
BarmarBarmar
420k35244344
add a comment |
3
If neither of the $_GET parameters is set, you never set $result to anything, so you'll get an error if you try to use it.
You should also use a prepared statement rather than substituting variables, to prevent SQL injection.
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$stmt = null;
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc and type= :type");
$stmt->execute(['loc' => $_GET['location'], 'type' => $_GET['type']]);
}
elseif (isset($_GET["location"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc");
$stmt->execute(['loc' => $_GET['location']]);
}
elseif (isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :type");
$stmt->execute(['type' => $_GET['type']]);
}
if ($stmt) {
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
} else {
$rows = ;
}
echo json_encode($rows);
?>
answered Mar 29 '18 at 20:54
BarmarBarmar
420k35244344
add a comment |
3
3
3
If neither of the $_GET parameters is set, you never set $result to anything, so you'll get an error if you try to use it.
You should also use a prepared statement rather than substituting variables, to prevent SQL injection.
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$stmt = null;
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc and type= :type");
$stmt->execute(['loc' => $_GET['location'], 'type' => $_GET['type']]);
}
elseif (isset($_GET["location"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc");
$stmt->execute(['loc' => $_GET['location']]);
}
elseif (isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :type");
$stmt->execute(['type' => $_GET['type']]);
}
if ($stmt) {
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
} else {
$rows = ;
}
echo json_encode($rows);
?>
answered Mar 29 '18 at 20:54
BarmarBarmar
420k35244344
If neither of the $_GET parameters is set, you never set $result to anything, so you'll get an error if you try to use it.
You should also use a prepared statement rather than substituting variables, to prevent SQL injection.
<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$stmt = null;
if(isset($_GET["location"]) && isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc and type= :type");
$stmt->execute(['loc' => $_GET['location'], 'type' => $_GET['type']]);
}
elseif (isset($_GET["location"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :loc");
$stmt->execute(['loc' => $_GET['location']]);
}
elseif (isset($_GET["type"]))
{
$stmt = $conn->prepare("Select * from resit_accommodation where location= :type");
$stmt->execute(['type' => $_GET['type']]);
}
if ($stmt) {
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
} else {
$rows = ;
}
echo json_encode($rows);
?>
answered Mar 29 '18 at 20:54
BarmarBarmar
420k35244344
answered Mar 29 '18 at 20:54
BarmarBarmar
420k35244344
answered Mar 29 '18 at 20:54
BarmarBarmar
420k35244344
answered Mar 29 '18 at 20:54
BarmarBarmar
420k35244344
420k35244344
add a comment |
add a comment |
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2
If none of the
ifconditions are true, you never set$result.– Barmar
Mar 29 '18 at 20:48
2
You also are open to SQL injections. You should parameterize the parameters you are passing in.
– chris85
Mar 29 '18 at 20:51
WARNING: When using PDO you should be using prepared statements with placeholder values and supply any user data as separate arguments. In this code you have potentially severe SQL injection bugs. Never use string interpolation or concatenation and instead use prepared statements and never put
$_POST,$_GETor any user data directly in your query. Refer to PHP The Right Way for general guidance and advice.– tadman
Mar 29 '18 at 21:13