隐藏表单并使用if语句

Question

I have looked in quite a few places and cannot find the a really good example of what I need. I have a button that when submitted needs to pop up a form. I have done this in the past using the input=hidden name. However, I am using trying to use if statements instead of just a variable. I will give you the code that I have in hopes that you can help find out what exactly I am doing wrong. You can see that I am indeed using the hidden name for some but I really don't know how to incorporate that into the if statements.

if($row[status] == 0) { print "<input type='image' src='images/rework-ticket.png' alt='Rework' value='Rework' name='button' style='height:50px; width:50px; '>"; }  
            elseif($row[status] == "Rework (In Progress)") { print "<img src='images/rework-ticket.png' alt='Rework' style='height:50px; width:50px; '>"; }


            print "<td align='center'>";
            print "<form method='post' action='test2.php'>";            
            print "<input type='image' src='images/rework-ticket.png' alt='Add Rework Ticket' value='Enter Employee ID' name='button' style='height:50px; width:50px; '>";
            print "<input type='hidden' name='proceed_to_rework' value='true'>";
            print "<input type='hidden' name='invoice_number' value='$row[invoice_number]'>";
            print "<input type='hidden' name='last_name' value='$row[last_name]'>";
            print "<input type='hidden' name='status' value='$status'>";
            print "Status:";
            print "<td><select name='status' size='1'>";
            if($status != NULL) { print "<option value='$status'>$status</option>"; }
            if($status != "Parts Prep (In Progress)") { echo "<option value='Parts Prep (In Progress)'>Parts Prep (In Progress)</option>"; }
            if($status != "Parts Prep (Complete)") { echo "<option value='Parts Prep (Complete)'>Parts Prep (Complete)</option>"; }     
            if($status != "Assembly (In Progress)") { echo "<option value='Assembly (In Progress)'>Assembly (In Progress)</option>"; }
            if($status != "Assembly (Complete)") { echo "<option value='Assembly (Complete)'>Assembly (Complete)</option>"; }       
            if($status != "Finish (In Progress)") { echo "<option value='Finish (In Progress)'>Finish (In Progress)</option>"; }
            if($status != "Finish (Complete)") { echo "<option value='Finish (Complete)'>Finish (Complete)</option>"; }     
            if($status != "Plumbing (In Progress)") { echo "<option value='Plumbing (In Progress)'>Plumbing (In Progress)</option>"; }
            if($status != "Plumbing (Complete)") { echo "<option value='Plumbing (Complete)'>Plumbing (Complete)</option>"; }
            print "</form>";
            print "</td>";

I really appreciate any comments on how I can not only fix this code but improve this question.

Answer 1

You can use the if statement as follows.

<?php if (condition): ?>

    <!-- HTML here -->
    <?php if (anotherCondition): ?>
        <!-- HTML here -->
    <?php endif; ?>
    <!-- HTML again -->

    <?php if (anotherCondition): ?>
        <!-- HTML here -->
    <?php endif; ?>
    <!-- HTML again -->

<?php endif; ?>

With this way, you don't have to use that much print statements.

Read the examples in php docs.