php - How to display default post with get method?
28
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"; }
Answer
Solution:
Asaph pointed out SQL injection. You should bind the parameter (google it), or at the minimum do this:
Answer
Solution:
If you want either
$_GET['productid']
or themax(productid)
when$_GET['productid']
is not set, you can use a ternary to change your sql queryso if
isset($_GET['productid'])
your query would bebut if not the default would be