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Hello, I need to create a webservice, but I have little experience with php and mysql, so I have some questions, I am using XAMPP and Mysql Workbench, in Workbench I did the connection test and said it worked, so maybe it is not a problem with Mysql. The webservice I am creating is to make the contents of the database available in a JSON, very simple indeed. I am basing myself on at that link, if anyone has any tutorial for this. I read that you may have to use PDO, it will be?
<?php
$host="XXXXX"; //replace with database hostname
$username="XXXXX"; //replace with database username
$password="XXXXX"; //replace with database password
$db_name="XXXXXX"; //replace with database name
$con=mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql = "select * from emp_info";
$result = mysql_query($sql);
$json = array();
if(mysql_num_rows($result)){
while($row=mysql_fetch_assoc($result)){
$json['emp_info'][]=$row;
}
}
mysql_close($con);
echo json_encode($json);
?>
The database I can create, as well as update the values, but I have this problem and I doubt if it is a problem with Mysql, with XAMPP, PHP or some configuration that remains to be done. If anyone can give me some direction to know what I should chase to resolve.
You haven’t described your problem, but you can see you’re using
mysql_*()
, this extension has been discontinued to learn more Why should we not use mysql type functions_*?– NoobSaibot
The problem is in relation to the functioning, I will try to use another function. But I think that’s precisely, thank you for the quick response.
– Marcos Max
Complementing what Wellingthon said, if you are using PHP7+ you will not be able to use
mysql_...
because in that version no longer exists, and you will even have to usemysqli_...
– Isac