viewing a column in mysql database using php script

Question

I'm trying to print a row from a table in mysql using a PHP script - here is the code but its not working - it shows Bad query apparently there is no issue:

$conn = mysqli_connect($servername, $username, $password, $db);
if (!$conn) {
 die("Connection failed: " . mysqli_connect_error());
}
else{
echo "Connected successfully";
}
$any = "show columns from help_category";
$chk = mysqli_query($any) or die("Bad Query: $sql");
echo"<table border='3'>";
echo"<tr><td>Type</td><tr>";
while($row = mysqli_fetch_assoc($chk))
{
echo"<tr><td>{$row['Type']}</td><tr>";
}
echo"</table>";

• zeehasham gt: this script just shows connected successfully and Bad Query after that, i cant find any bad query logs in the mysql log files. Can you confirm where shall i check them ? im using debian 9
• Vérace: First - run the query "raw" from the mysql CLI (Command Line Interface which, funnily enough, is called mysql!). From there you will have a meaningful (i.e. direct from the server's mouth) error message - when your SQL works from mysql, then "translate" it into PHP! This should be your default method of working!
• zeehasham gt: i have run these queries in mysql already and they are working there. it seems that its any issue while i embed it in php.

Answer 1

I have checked your code and i found out that you have issue with your query execution on php function mysqli_query

mysqli_query requires two parameter first one is connection object and second one is query variable to execute a query.

Below is exact code:-

$conn = mysqli_connect($servername, $username, $password, $db);
if (!$conn) {
 die("Connection failed: " . mysqli_connect_error());
}
else{
echo "Connected successfully";
}
$any = "show columns from transaction";
$chk = mysqli_query($conn, $any) or die("Bad Query: $sql");
echo"<table border='3'>";
echo"<tr><td>Type</td><tr>";
while($row = mysqli_fetch_assoc($chk))
{
echo"<tr><td>{$row['Type']}</td><tr>";
}
echo"</table>";