update MySQL in PHP - php

Question

I want to update MySQL database in PHP. My problem is that I have 3 tables. An employee, a company, and an employee_company (it has the another 2 table's id). And I want to update the employee table if I change the employee's name and the employee_company table if I add a company to the employee. I tried to solve this problem, but it didn't update the tables. How can I do this?
index.php
<form method="post" id="insert_form">
<label>Enter Employee Name</label>
<input type="text" name="name" id="name" class="form-control" />
<br />
<label>Enter Employee Address</label>
<textarea name="address" id="address" class="form-control"></textarea>
<br />
<label>Company:</label>
<select name="company" id="company" class="form-control">
<?php
$query2 = "SELECT * FROM company ORDER BY company_id";
$result2 = mysqli_query($connect, $query2);
while($row2= mysqli_fetch_array($result2)){
?>
<option value="<?php echo $row2['company_id'];?>"><?php echo $row2['company'];?></option>
<?php
}
?>
</select>
<br/>
<input type="hidden" name="employee_id" id="employee_id" />
<input type="submit" name="insert" id="insert" value="Insert" class="btn btn-success" />
</form>
<script>
$(document).on('click', '.edit_data', function(){
var employee_id = $(this).attr("id");
$.ajax({
url:"fetch.php",
method:"POST",
data:{employee_id:employee_id},
dataType:"json",
success:function(data){
$('#name').val(data.name);
$('#address').val(data.address);
$('#company').val(data.company);
$('#employee_id').val(data.id);
$('#insert').val("Update");
$('#add_data_Modal').modal('show');
}
});
});
$('#insert_form').on("submit", function(){
else
{
$.ajax({
url:"insert.php",
method:"POST",
data:$('#insert_form').serialize(),
beforeSend:function(){
$('#insert').val("Inserting");
},
success:function(data){
$('#insert_form')[0].reset();
$('#add_data_Modal').modal('hide');
$('#employee_table').html(data);
}
});
}
});
</script>
insert.php
$name = mysqli_real_escape_string($connect, $_POST["name"]);
$company = mysqli_real_escape_string($connect, $_POST["company"]);
$id=$_POST["employee_id"];
if($_POST["employee_id"] != '')
{
$query = "
UPDATE employee, employee_company
SET employee_company.employee_id='$id',
employee_company.company_id='$company',
employee.name='$name',
WHERE employee.employee_id='".$_POST["employee_id"]."'";
$message = 'Data Updated';
}

Answer 1

You were missing mysqli_query() function which actually runs the query. Your$query variable is just a string variable until you don't put it into mysqli_query() function
Replace your code with this piece of code
if($_POST["employee_id"] != '')
{
$query = "
UPDATE employee, employee_company
SET employee_company.employee_id='$id',
employee_company.company_id='$company',
employee.name='$name'
WHERE employee.employee_id='".$_POST["employee_id"]."'";
if(mysqli_query($connect,$query))
$message = 'Data Updated';
else
echo mysqli_error($connect);
}
If there is any query then you can ask me freely!

Answer 2

There is some mistake in query, using additional comma(,) before where condition. So remove that. Check the updated code below:
if ($_POST["employee_id"] != '') {
$query = "UPDATE employee, employee_company SET employee_company.employee_id='$id', employee_company.company_id='$company', employee.name='$name' WHERE employee.employee_id='" . $_POST["employee_id"] . "'";
mysqli_query($connect,$query);
$message = 'Data Updated';
}
Note: Also I recommend you to sanitize all inputs before using in query.