php dynamic list help

[MBrown]

I'm trying to make a dynamic list that pulls a pk from a database, but I am running into some stupid parse errors and I cannot see what the issue is. Can someone with some fresh eyes check this out? Thanks!

This is the error: Parse error: syntax error, unexpected '"'>"' (T_CONSTANT_ENCAPSED_STRING)

this is the function

function serialDropdown($listQuery, $con)
	{
	 $list = mysql_query($listQuery, $con);
	 $menu = "<select name = 'serialNumberList'>";
	 while ($result = mysql_fetch_array($list))
	 {
		// The line below is the one that keeps giving me errors!
		$menu .= "<option value=' ".$result['serialNumber'] "'>" .$result['serialNumber'] ." </option>";
	 }
		$menu .= "</select>";
		return $menu;
         }

UPDATE!!!!!!!

Ok. I'm trying a different approach now, but for some reason it is not displaying my first result. For example in my table I have 7 rows but its only displaying 2 through 7 in my list. I cannot see where my issue is. Anyone else see it?

$listRows = mysqli_num_rows($listQuery);
	echo "<br /> <br /> <br />";
	echo "<select name = 'serialNumberlist'>";
	for ($y = 0; $y < $listRows; ++$y)
	{
		$listRows = mysqli_fetch_row($listQuery);
		for ($x = 0; $x < 1; $x++)
		{
			echo "<option value = '$listRows[$x]'> $listRows[$x] </option>";
		}
	}
        echo "</select>";

Thanks in advance!

 

[Markbnj]

Not a php guy, but going out on a limb to suggest you take a really careful look at the order and number of single vs. double quotes.

 

[MBrown]

Not a php guy, but going out on a limb to suggest you take a really careful look at the order and number of single vs. double quotes.

Double check and double checked. I was using notepad++ for this but I installed phpstorm 30 day trial and its saying that it needs a semicolin to end the line of code but that is just not correct.

 

[DaveSimmons]

Missing a . (period) before the " for the string concatenation

.$result['serialNumber'] "'>"

That's why it's saying you need a ; there.

 

[MBrown]

Ok. I'm trying a different approach now, but for some reason it is not displaying my first result. For example in my table I have 7 rows but its only displaying 2 through 7 in my list. I cannot see where my issue is. Anyone else see it?

$listRows = mysqli_num_rows($listQuery);
	echo "<br /> <br /> <br />";
	echo "<select name = 'serialNumberlist'>";
	for ($y = 0; $y < $listRows; ++$y)
	{
		$listRows = mysqli_fetch_row($listQuery);
		for ($x = 0; $x < 1; $x++)
		{
			echo "<option value = '$listRows[$x]'> $listRows[$x] </option>";
		}
	}
        echo "</select>";

Thanks in advance!

 

[sathyan]

try this:
$listRows = mysqli_num_rows($listQuery);
echo "<br /> <br /> <br />";
echo "<select name = 'serialNumberlist'>";
for ($y = 0; $y < $listRows; ++$y)
{
$listRows = mysqli_fetch_row($listQuery);
for ($x = 0; $x < 1; $x++)
{
echo "<option value = ".$listRows[$x].">". $listRows[$x]."</option>"; // changed quote type and add concatenation dots
}
}
echo "</select>";

 

[DaveSimmons]

Why are you using $listRows for both the count and the fields for each fetched row?

try using $row inside the loop, or $listcount for the count of rows.

 

[Albatross]

When you call mysqli_fetch_row it increments the internal pointer so that`s why when you echo the dropdown it skips one.

 

[MBrown]

Why are you using $listRows for both the count and the fields for each fetched row?

try using $row inside the loop, or $listcount for the count of rows.

tried this but it didn't work. I used this same loop set up to display a table of data and it works perfectly. I figured I could implement this into my drop down list. Also it seems that it is displaying the correct number of options, but the first option always says its 2 and the last option is always blank.