MySQL contains variable - php

Question

I have an array in PHP that is looping through a set of names (and a corresponding quantity). I would like to print the ones found in a MYSQL database table (to which I've succesfully connected). I'm currently using the code:
foreach ($arr as $name => $quan) {
$query = "SELECT * FROM table WHERE name='$name'";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) > 0) {
$row = mysql_fetch_array($result);
echo $quan." ".$row['name']. "\n";
}
}
For some reason, this only prints the last quantity and name in the array. Help?
For example, if the array has key-value pairs of {A-4, B-2, C-3}, and table contains {A, B, D} as names ... it'll only print "2 B".

Answer 1

Change the code to the following:
foreach ($arr as $name => $quan) {
$query = "SELECT * FROM table WHERE name='$name'";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) > 0) {
while($row = mysql_fetch_array($result)) {
echo $quan." ".$row['name']. "\n";
}
}
}
You have to loop through the result. By the way, stop using mysql_query()! use MySQLi or PDO instead ( and be careful of SQL Injection ; you can use mysqli_real_escape_string() to handle input parameters ). For MySQLi implementation , here it is :
foreach ($arr as $name => $quan) {
$query = "SELECT * FROM table WHERE name='$name'";
$result = mysqli_query($query) or die(mysqli_error());
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result)) {
echo $quan." ".$row['name']. PHP_EOL;
}
}
}
And rather "\n", I prefer using PHP_EOL ( as shown above )
And as the comment suggests, the SQL statement can be executed once, as follow:
$flipped_array = array_flip($arr); // flip the array to make "name" as values"
for($i = 0; $i < count($flipped_array); $i++) {
$flipped_array[$i] = '\'' . $flipped_array[$i] . '\''; // add surrounding single quotes
}
$name_list = implode(',', $arr);
$query = "SELECT * FROM table WHERE name IN ($name_list)";
// ... omit the followings
e.g. in $arr contains "peter", "mary", "ken", the Query will be:
SELECT * FROM table WHERE name IN ('peter','mary','ken')
sidenote: but I don't understand your query. You only obtain the name back from the query? You can check number of rows, or you can even group by name, such as:
SELECT name, COUNT(*) AS cnt FROM table GROUP BY name ORDER BY name
to get what you want.
UPDATE (again): based on the comment of OP, here is the solution :
$flipped_array = array_flip($arr); // flip the array to make "name" as values"
for($i = 0; $i < count($flipped_array); $i++) {
$flipped_array[$i] = '\'' . $flipped_array[$i] . '\''; // add surrounding single quotes
}
$name_list = implode(',', $arr);
$query = "SELECT name, COUNT(*) AS cnt FROM table WHERE name IN ($name_list) GROUP BY name HAVING COUNT(*) > 0 ORDER BY name";
$result = mysqli_query($query) or die(mysqli_error());
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result)) {
echo $quan." ".$row['name']. ": " . $row['cnt'] . PHP_EOL;
}
}
The above query will show the name appearing in the table only. Names not in table will not be shown. Now full codes ( be cautious of SQL Injection , again )