Check if match the url parameter if correct and display the data

Question

Im newbie in programmin I want Check if the 2 parameter are correct and display the data

www.site.com/?param1=123&param2=444 <--- check if param1 and param2 are match and display the account.

On the database it has two same param2 and but different param1.

param1 | param2

123 444 First 122 444 Second

I want to display only the first

$param1 = $_GET['param1'];
$param2 = $_GET['param2'];

$result = mysqli_query($con,"SELECT * FROM tests");

if (mysqli_num_rows($result) > 0) {

while($row = mysqli_fetch_array($result)) {

}

}


if ( $param1 ==($param2)){

    echo "Display the data all the of first";

}

else{

    echo "incorrect parameter;";
}

This is an example of how to deal with parameters: https://stackoverflow.com/questions/728229/parameters-in-mysqli — Luuk, May 03 '20 at 13:20

score 0 · Answer 1 · answered May 03 '20 at 13:26

0

Try mysqli_fetch_all instead of mysqli_fetch_array in cycle: $rows = mysqli_fetch_all($result, MYSQLI_ASSOC);.

And then choose needed row to show (if it exists): if (isset($rows[0])) { // your code to show }.

answered May 03 '20 at 13:26

Slava

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Check if match the url parameter if correct and display the data

1 Answers1