PHP Update content chosen in a drop down list in a form on edit
I am trying design a form which involves a drop down list. I would like to edit the form by changing the value of the variables in the form. I have retrieved all the values from mysql DB, but when I click the submit button without changing the value in the dropdown list. The value will be set to null.
if(isset($_POST['update'])){
$update_id = $_GET['edit_form'];
$post_title1 = $_POST['title'];
$post_date1 = date("Y-m-d");
$post_author1 = $_POST['author'];
$post_keywords1 = $_POST['keywords'];
$post_type1 = $_POST['category']; // This is the content I want to update but was set to null, the rest is fine.
$post_content1 = $_POST['content'];
$post_image1 = $_FILES['image']['name'];
$image_tmp = $_FILES['image']['tmp_name'];
}}
This is my dropdown list:
<html>
<select name="category" class="err">`enter code here`
<option value=""><?php echo $post_type;?></option>
<option value="Class">Class</option>
<option value="Facilities">Facilities</option>
<option value="Services">Services</option>
</select>
What I want is when I click the submit button without changing the content of the post, the $post_type1 will not be set to null, but keep its original value from DB.
Thanks, hope I have stated my question clearly:)
Here's the code that builds the db query :
$post_image1 = $_FILES['image']['name'];
move_uploaded_file($image_tmp,"../images/$post_image1");
$update_query = "UPDATE post SET post_title='$post_title1', post_date='$post_date1', post_author='$post_author1', post_image='$post_image1', post_keywords='$post_keywords1',post_type='$post_type1',
post_content='$post_content1' WHERE post_id ='$update_id'";
In HTML you should set a default value for the dropdown.
You write a select element with options. The current database value's option element should have the boolean selected attribute set.
How can I set the default value for an HTML <select> element?
<select name="category" id="category">
<option></option>
<option></option>
<option selected="selected">
</option>
<option></option>
</select>
This way $post_type1 will hold the selected value if you click submit without modifying the form.
Now for the image. you have to test values to build the right mySQL query:
$update_query = "UPDATE post SET post_title='$post_title1', post_date='$post_date1', post_author='$post_author1'";
$post_image1 = isset($_FILES['image']) ? $_FILES['image']['name'] : false;
if($post_image1 !== false){
if(move_uploaded_file($image_tmp,"../images/$post_image1")){
//file successfully moved
$update_query .= " , post_image='$post_image1'";
}
}
$update_query .= ", post_keywords='$post_keywords1',post_type='$post_type1',
post_content='$post_content1' WHERE post_id ='$update_id'";
You get the gist of it!?
Best way to do it would be have a default value of the drop down to be like -1 .eg
HTML
<select name="category">
<option id="-1">Please Select</option>
<option id="1">Category 1</option>
<option id="2">Category 2</option>
</select>
PHP
<?php
$post_type1 = ( 0 < ( int ) $_POST[ 'category' ] ) ? $_POST[ 'category' ] : null;
Edit : Sorry, misread the question
This is probably more what you want :
PHP
<?php
$mysqli = new mysqli( "localhost", "xxx", "xxx", "xxx" );
$query = "SELECT * FROM xxx";
$options = array();
$category = ( int ) $_POST[ 'category' ];
$query = $mysqli->query( $query );
if( FALSE !== $query )
{
while( $row = $mysliResults->fetch_assoc() )
{
$selected = "";
if( $category === ( int ) $row['id'] )
$selected = " selected="selected"";
$options []= "<option id="{$row['id']}"{$selected}>{$row['name']}</option>":
}
}
echo "<select name="category">", implode( "nt", $options ), "</select>";
Hope that helps
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