Populating a form on another page with a hyperlink

[jaady]

I am trying to populate a form with the data collected from another page.

I run a select * query in a page, format it into a table and added a hyperlink at the end of the $row. The idea is to pull this record out and bring it into my update form so I can update information in the record.

the problem that I am having is no matter what $row I select and click the hyperlink I always end up getting the last record of my database and not the record that I am trying to select. I’ve been playing around for a while and not getting any closer to figure it out. any idea whats happening?

update

the receiving input field in my table on the other page:

[Drummin]

You would need to show some code. The $row[‘field’] would need to be in the WHILE loop.

[jaady]

 //query to populate employee form//
		$query = "SELECT * FROM employee ORDER BY firstname ASC";
		$result = mysqli_query($dbc, $query);
		
	

	echo'<table class="tblviewemp" >';
		
		echo '<thead>';
			echo '<tr class="tblviewemp">';
			echo'<th class="dataH" id="emppicthumb">Employee Pic </th>';
			echo'<th class="dataH"> First Name</th>';
			echo'<th class="dataH">Last Name </th>';
			echo'<th class="dataH"> ID</th>';
			echo'<th class="dataH"> Update</th>';
		echo '</thead>';
		
	while ($row = mysqli_fetch_array($result)) {
		  // Display the employee data
		
		echo '<tr>';
			//echo '<td class="thumb"><img  src="' . EMP_UPLOADPATH . $row['employeepic'] . '" alt="Employee image" /></td>';
			echo '<td class="data">' . $row['firstname'] . '</td>';
			echo '<td class="data">' . $row['lastname'] . '</td>';
			echo '<td class="data">' . $row['employeeid'] . '</td>';
			echo '<td><a class="update" href="update_employee.php?employeeid=' . 
			$row['employeeid'] . '&amp;firstname='. $row['firstname'] . '&amp;lastname=' . $row['lastname'] .'">update</a></td>';
	}
		echo '</tr>';
	echo'</table>';

and the receiving end is:

                 <form>
                  <label for="employeeid" class="label" >Employee ID:</label>                
                  <input name="employeeid" type="text" required class="input" id="employeeid" value="<?php echo $_GET['employeeid']; ?>"/>
                  
                  <label for="firstname" class="label" >First Name:</label>                
                  <input name="firstname" type="text" required class="input" id="firstname" value="<?php echo $_GET['firstname']; ?>"/>
                  
                  <label for="lastname" class="label" >Last Name:</label>                
                  <input name="lastname" type="text" required class="input" id="lastname" value="<?php echo $_GET['lastname']; ?>" />
</form>

[jaady]

ok this didn’t print out properly… let me try again

the sending script is:

 //query to populate employee form//
		$query = "SELECT * FROM employee ORDER BY firstname ASC";
		$result = mysqli_query($dbc, $query);
		
	

	echo'<table class="tblviewemp" >';
		
		echo '<thead>';
			echo '<tr class="tblviewemp">';
			echo'<th class="dataH" id="emppicthumb">Employee Pic </th>';
			echo'<th class="dataH"> First Name</th>';
			echo'<th class="dataH">Last Name </th>';
			echo'<th class="dataH"> ID</th>';
			echo'<th class="dataH"> Update</th>';
		echo '</thead>';
		
	while ($row = mysqli_fetch_array($result)) {
		  // Display the employee data
		
		echo '<tr>';
			//echo '<td class="thumb"><img  src="' . EMP_UPLOADPATH . $row['employeepic'] . '" alt="Employee image" /></td>';
			echo '<td class="data">' . $row['firstname'] . '</td>';
			echo '<td class="data">' . $row['lastname'] . '</td>';
			echo '<td class="data">' . $row['employeeid'] . '</td>';
			echo '<td><a class="update" href="update_employee.php?employeeid=' . 
			$row['employeeid'] . '&amp;firstname='. $row['firstname'] . '&amp;lastname=' . $row['lastname'] .'">update</a></td>';
	}
		echo '</tr>';
	echo'</table>';

and the receiving form is:

<form>
<label for="employeeid" class="label" >Employee ID:</label>                
                  <input name="employeeid" type="text" required class="input" id="employeeid" value="<?php echo $_GET['employeeid']; ?>"/>
                  
                  <label for="firstname" class="label" >First Name:</label>                
                  <input name="firstname" type="text" required class="input" id="firstname" value="<?php echo $_GET['firstname']; ?>"/>
                  
                  <label for="lastname" class="label" >Last Name:</label>                
                  <input name="lastname" type="text" required class="input" id="lastname" value="<?php echo $_GET['lastname']; ?>" />
</form>

[jaady]

for some reason the first couple of line don’t want to print in here: trying again with the missing lines:

 <label for="employeeid" class="label" >Employee ID:</label>                
                  <input name="employeeid" type="text" required class="input" id="employeeid" value="<?php echo $_GET['employeeid']; ?>"/>

[jaady]

Well the site doesn’t want to print my first two line which are the same: label and input for employeeid

[Drummin]

When viewing source, do you see correct info? Have you tried moving class=“update” after href=? I have not see <a href broken like that before.

[jaady]

You my friend are getting too good at this… That was the problem. the class between “a” and “href” caused the problem. its loading perfectly now. thanks a bunch. Oh! and I hate you hahahaha I been working on this and reading for like 4 hours and it took you 1 look to figure it out. you are quite skilled. thank you very much.

[TechnoBear]

@jaady

Well the site doesn’t want to print my first two line

I’ve fixed the posts now.

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