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POInsert POST data to table and POST array to seperate table
 

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    <p>I am trying to submit data from a form to two separate tables. </p> <p><strong>Here's the error:</strong> It inserts into table 1 fine but table2 array data goes into the database as "<strong>Array</strong>".</p> <p>Here are my fields going into table1:</p> <pre><code> $start = $_POST['start']; $end = $_POST['end']; $customer = $_POST['customer']; $manufacturer = $_POST['manufacturer']; $rep = $_POST['rep']; $notes = $_POST['notes']; </code></pre> <p>My array fields going into table2:</p> <pre><code> item[] description[] pack[] </code></pre> <p>Any help is appreciated. Below is the code I have developed thus far:</p> <pre><code> if ($start == '' || $end == '') { $error = '&lt;div class="alert alert-error"&gt; &lt;a class="close" data-dismiss="alert"&gt;×&lt;/a&gt; &lt;strong&gt;Error!&lt;/strong&gt; Please fill in all required fields! &lt;/div&gt;'; } else { $sql = "SELECT COALESCE(MAX(GroupID), 0) + 1 AS newGroupID FROM table1"; try { $stmt = $db-&gt;prepare($sql); $stmt-&gt;execute(); } catch(PDOException $ex) { die("Failed to run query: " . $ex-&gt;getMessage()); } $rows = $stmt-&gt;fetchAll(); foreach($rows as $row) { $groupID = $row['newGroupID']; } $mysqli = new mysqli("localhost", "user", "pw", "mydb"); if (mysqli_connect_errno()) { die(mysqli_connect_error()); } $start = $_POST['start']; $end = $_POST['end']; $customer = $_POST['customer']; $manufacturer = $_POST['manufacturer']; $rep = $_POST['rep']; $notes = $_POST['notes']; if ($stmt = $mysqli-&gt;prepare("INSERT table1 (GroupID, start, end, customer, manufacturer, rep, notes) VALUES (?, ?, ?, ?, ?, ?, ?)")) { $stmt-&gt;bind_param("issssss", $groupID, $start, $end, $customer, $manufacturer, $rep, $notes); $stmt-&gt;execute(); $stmt-&gt;close(); } else { echo "ERROR: Could not prepare SQL statement 1."; } $mysqli-&gt;error; $mysqli-&gt;close(); $success = "&lt;div class='alert alert-success'&gt;New agreement added.&lt;/div&gt;"; $mysqli = new mysqli("localhost", "user", "pw", "mydb"); if (mysqli_connect_errno()) { die(mysqli_connect_error()); } if ($stmt = $mysqli-&gt;prepare("INSERT table2 (GroupID, item_number, item_description, pack) VALUES (?, ?, ?, ?)")) { foreach($_POST['item'] as $i =&gt; $item) { $item = $_POST['item']; $description = $_POST['description']; $pack = $_POST['pack']; } $stmt-&gt;bind_param("isss", $GroupID, $item, $description, $pack); $stmt-&gt;execute(); $stmt-&gt;close(); } else { echo "ERROR: Could not prepare SQL statement 2."; } $mysqli-&gt;error; $mysqli-&gt;close(); $success = "&lt;div class='alert alert-success'&gt;New agreement items added!&lt;/div&gt;"; } } </code></pre>
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    1. COwhat is not working? are you getting any error message ?
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      1. USMaximus2012
    2. CO@Maximus2012 Just updated my post (see top). Table1 inserts fine but table2 array data inserts as "Array" rather than the actual data.
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    3. COAre you using both `PDO` (1 connection) AND `mysqli` (2 separate connections)
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      1. USSean
 

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