html select inside table

[phpclown]

i have a table that is built off a select statement pulling records from mysql. the table has the following fields.

 

id,transaction_date,description,check_number,amount,transaction_category

 

I import a csv that has all the info but the transaction_category. I want to have a select box that is created from a separate query in each row of the html table. The idea is i can quickly assign a category to a transaction using the drop down and save my changes. I am stuck on to areas.

 

1. The select box is off by one column in my table

2. Only the first select box has data.

 

I get a correct looking table with all the data and the select dropdown on every row.

echo '<table border="1" cellpadding="5" cellspacing="5" class="db-table">';
		echo '<tr><th>ID</th><th>Date</th><th>Description</th><th>Check Number</th><th>Amount<th>Category</th></tr>';
		while($row = $sql_select->fetch())
                {
			echo '<tr>';
			foreach($row as $key=>$value)
                        {
				echo '<td>',$value,'</td>';
                                
                         
			}
                        
                               echo "<td>";
                                echo "<select name=\"category\">";
                            while($row2 = mysql_fetch_array($results_cate)) 
                            {        
                                echo "<option value='".$row2['category']."'>".$row2['category']."</option>";
                                
                            }
                            echo "</select>";
                            echo "<td>";
                        
                        echo "</tr>";
			//echo '</tr>';
		}
		echo '</table><br />';

[jcbones]

Are all the select boxes the same? Except the name of course.

[davidannis]

You need to allow multiple values for category so change this

echo "<select name=\"category\">";

to

echo "<select name=\"category[]\">";

You will then get an array $_POST['category'][] with all of your category values.

[davidannis]

FWIW you can also code it like this

echo "<select name=\"category[$value]\">";

and then you get an array $_POST['category'] that looks like this ('myvalue1'=>'mycategory', 'myvalue2'=>'mycat2', ...)

[phpclown]

well while i was trying the suggestions above i realized i have another issue. I am converting over to PDO and i know nothing about it. Looks like i can't have two querys. Here is what i have. It gives me a php parser error on the second query. I can't test the solution for the html select untill i get this fix i guess. I can post a new topic if thats what i need to do.

 try
 {
  $DBH = new PDO("mysql:host=localhost;dbname=xxxxxx", xxxx, xxxx);
 }
 catch(PDOException $e) {  
    echo $e->getMessage();
 }
 $sql_select = $DBH->query('SELECT * FROM `transaction`');

# setting the fetch mode  
$sql_select->setFetchMode(PDO::FETCH_OBJ);  
$results = mysql_query($sql_select);

//sql_select_cate = $DBH->query('select * from `categories`');

# setting the fetch mode  
sql_select_cate->setFetchMode(PDO::FETCH_ASSOC);
$results_cate = mysql_query($sql_select_cate);

[Jessica]

You should uncomment the line that has the query.

If you have an error, post it here.

[phpclown]

here you go.

 

[14-Apr-2013 14:33:09] PHP Parse error:  syntax error, unexpected '=' in /Applications/MAMP/htdocs/finance/assigncate.php on line 25

 

this is line 25

 

sql_select_cate = $DBH->query('select * from `categories`');

[jcbones]

$ <- you missed it.

[phpclown]

wow thats embarassing. 

 

Now back to the orginal post.  I added the [ ] to the echo "<select name=\"category[]\">";

 

I still get one drop down with data and it off by one colum.

[Barand]

When you create the first selection you loop through all the rows until none are left. When you try to create the next there are none to read.

 

Create the options first and store in a string var then just output the string each time.

[phpclown]

Thanks Barand that works. I also realized that my problem with the select box being off by a row was do to returning an extra column in my sql query. I am going to mark this solved. Thanks everyone for the help.