PHP Insert table into database - php

Question

I have a database table that I created from a .csv file. I used php to edit the content to match a new table on my database. I now need to INSERT that new table into the new database. I added form tags around the table and a submit button, and when I submit it I only get 1 or 2 random rows of the table. I am guessing that I might need a while or a for loop, but I cant figure out how to write it. Any assistance would be appreciated.
This is the code to post to the db
if(isset($_POST['submit'])){
$mr = $_POST['mr'];
$mrs = $_POST['mrs'];
$last_name = $_POST['last_name'];
$marital_status = $_POST['marital_status'];
$query = "INSERT INTO import_template(first_name, last_name, marital_status) VALUES('{$mr}', '{$last_name}', '{$marital_status}')";
$result = mysqli_query($connection, $query);
if(!$result){
die("QUERY FAILED " . mysqli_error($connection));
}
if(!$connection){
echo "We are NOT connected";
}
$query = "INSERT INTO import_template(first_name, last_name, marital_status) VALUES('{$mrs}', '{$last_name}', '{$marital_status}')";
$result = mysqli_query($connection, $query);
if(!$result){
die("QUERY FAILED " . mysqli_error($connection));
}
if(!$connection){
echo "We are NOT connected";
}
}
This is what the dynamic table looks like I only added the input to the first few to test, I haven't done the rest yet.
echo "<tr>";
echo "<td><input type='text' class='form-control' name='mr' value='$mr'></td>";
echo "<td><input type='text' class='form-control' name='last_name' value='$last_name'></td>";
echo "<td><input type='text' class='form-control' name='marital_status' value='$marital_status'></td>";
echo "<td>{$birthday}</td>";
echo "<td>Adult</td>";
echo "<td>{$anniversary}</td>";
echo "<td>{$mr_cell_phone}</td>";
echo "<td>{$home_phone}</td>";
echo "<td>{$mr_email}</td>";
echo "<td>{$street}</td>";
echo "<td>{$city}</td>";
echo "<td>{$state}</td>";
echo "<td>{$zip}</td>";
echo "<td></td>";
echo "</tr>";
echo "<tr>";
echo "<td><input type='text' class='form-control' name='mrs' value='$mrs'></td>";
echo "<td><input type='text' class='form-control' name='last_name' value='$last_name'></td>";
echo "<td><input type='text' class='form-control' name='marital_status' value='$marital_status'></td>";
echo "<td></td>";
echo "<td>Adult</td>";
echo "<td>{$anniversary}</td>";
echo "<td>{$mrs_cell_phone}</td>";
echo "<td></td>";
echo "<td>{$mrs_email}</td>";
echo "<td></td>";
echo "<td></td>";
echo "<td></td>";
echo "<td></td>";
echo "<td></td>";
echo "</tr>";

Answer 1

You can do like this. I have taken three fields such as task,agent and note.You can add as per your table rows:
main.php
<form id="first" action="insert_values.php" method="post">
<table>
<tr>
<td>Task</td>
<td>
<input type="text" name="task">
</td>
</tr>
<tr>
<td>Agent</td>
<td>
<input type="text" name="agent">
</td>
</tr>
<tr>
<td>Note</td>
<td><input type="text" name="Note"></td>
</tr>
</table>
<input type="submit" value="save" name="save">
</form>
insert_values.php
<?php
error_reporting( error_reporting() & ~E_NOTICE );
session_start();
include 'connect.php';
$task=$_POST['task'];
$agent=$_POST['agent'];
$Note=$_POST['Note'];
$result=mysql_query("insert into table_name(task,agent,t_note) values('$task','$agent','$Note')");
if($result)
{
$msg="Insert Successfully!!";
echo "<script>alert('Insert Successfully');document.location='main.php'</script>";
?>
<?php
}
else
{
$errmsg=" Insert not success!!";
echo "<script>alert('Insert Not Successfully');document.location='main.php'</script>";
}
?>