Having error when inserting my item name to my phpmyadmin table - php

Question

SO i want to insert my item id and together with my item name into a distribution_item,it only insert item id but not the name because of the array
i use item[$i] whenever there is a item in the table row, so it get the data and insert.
#$recipient = $_POST['recipient'];
#$address = $_POST['address'];
#$contact = $_POST['contact'];
#$date = $_POST['in_date'];
#$itemID = $_POST['id'];
#$remark = $_POST['remark'];
#$spec_remark = $_POST['spec_remark'];
$itemBalance = $_POST["count"];
$count = count($itemID);
// authentication to the database
$servername = "localhost";
$username = "root";
$password = "";
$dbName = "hopeplace";
//Create connection
$Conndb = mysqli_connect($servername, $username, $password, $dbName);
// Check connection
if (!$Conndb) {
die("Connection failed: " . $Conndb->connect_error);
}
else {
// select database
mysqli_select_db($Conndb, $dbName);
$full_name = "SELECT * FROM recipient WHERE `FULL_NAME` = '$recipient'";
$result = mysqli_query($Conndb, $full_name);
$rec = mysqli_fetch_array($result);
$recipient_id = $rec['HP_ID'];
$item_id = "SELECT ITEM_NAME FROM inventory WHERE ITEM_ID = '$itemID'";
$result2 = mysqli_query($Conndb, $item_id);
$rec2 = mysqli_fetch_array($result2);
$item_name = $rec2["ITEM_NAME"];
$string = implode(',',$item_id);
$sql = "SHOW TABLE STATUS WHERE `Name` = 'distribution';";
$result = mysqli_query($Conndb, $sql);
$data = mysqli_fetch_assoc($result);
$DISTRIBUTION_ID = $data['Auto_increment'];
// Add into distribution table
$sql = "INSERT INTO distribution(DISTRIBUTION_ID,HP_ID,FULL_NAME, ADDRESS, CONTACT, DISTRIBUTION_DATE, SPEC_REMARK) VALUES ('$DISTRIBUTION_ID','$recipient_id','$recipient', '$address', '$contact', '$date', '$spec_remark')";
if (mysqli_query($Conndb, $sql)) {
//Add item into distribution_item table
$item_count = 0;
for ($i=0; $i<$count; $i++){
$sql = "INSERT INTO distribution_item (DISTRIBUTION_ID, ITEM_ID,ITEM_NAME,OUT_QUANTITY,REMARK) VALUES ('$DISTRIBUTION_ID', '$itemID[$i]','$string[$i]','$itemBalance[$i]', '$remark[$i]')";
if (mysqli_query($Conndb, $sql)){
$out = "UPDATE inventory set QUANTITY = QUANTITY - '$itemBalance[$i]' where ITEM_ID= '$itemID[$i]'";
mysqli_query($Conndb, $out);
//echo "<p>Item $itemID[$i] has been added to $DISTRIBUTION_ID</p>";
$item_count++;
} else {
echo "Error: $sql <br />" . mysqli_error($Conndb);
}
}
if ($item_count == $count){
echo "<div>
<script>
window.alert('Record added successfully!');
</script>
</div>";
}
} else {
echo "Error: $sql <br />" . mysqli_error($Conndb);
}
}
mysqli_close($Conndb);
?>
it pop out an error like this
Notice: Array to string conversion in C:\xampp\htdocs\hopeplace\distribution\add_distribution.php on line 55
Warning: implode(): Invalid arguments passed in C:\xampp\htdocs\hopeplace\distribution\add_distribution.php on line 56
it seems like i have to convert my array to string so i can pass the value into table. my database table is something like this
DISTRIBUTION_ID | ITEM_NAME | ITEM_NAME|
1 1 APPLE
1 2 ORANGE

Answer 1

The problem here is that $string[$i] cannot be found because no $string array exists, since $string = implode(', ',$item_id) can only work if $item_id is an array.
In the question, $item_id = "SELECT ITEM_NAME FROM inventory WHERE ITEM_ID = '$itemID'"; gives $item_id as a string. That is why the warning error indicates that the implode function cannot work. The notice error shows that you cannot get an array element from the $string variable, and this is because that variable is not an array.
To get $string as an array, you would have to correct the following code:
$item_id = "SELECT ITEM_NAME FROM inventory WHERE ITEM_ID = '$itemID'";
$result2 = mysqli_query($Conndb, $item_id);
$rec2 = mysqli_fetch_array($result2);
$item_name = $rec2["ITEM_NAME"];
$string = implode(',',$item_id);
For example, the above code could be changed to the following:
$string = [];
$item_id = "SELECT ITEM_NAME FROM inventory WHERE ITEM_ID = '$itemID'";
$result2 = mysqli_query($Conndb, $item_id);
while($rec2 = mysqli_fetch_array($result2){
$string[] = $rec2["ITEM_NAME"];
}
Then in the sql query for inserting the item name you would first define the counting variable as $count = count($string);