Page hangs on multi command mysqli php - php

Question

So I'm pretty new to PHP and I'm mostly just struggling my way through it but I've hit a roadblock that I just can't get over.
My goal is to run a few commands to clear and update a standings table and then display the data in a table. I've got the table to display pretty much no problem but when I add the code in to update the table the page stops before it even displays the table and it's just a blank page, it wont even run any of the HTML.
Here's the PHP, I can provide the rest of the page code if needed.
http://pastebin.com/GZ3KcNXK
<?php
$con = mysqli_connect("localhost","uakronhv_admin","uakhvzdbadmin01","uakronhv_game1");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "DELETE FROM standings WHERE 1=1; ";
$sql. = "INSERT INTO standings (`pId`,`pName`,`pStarve`) SELECT p.pId, concat(p.pFirst," ",p.pLast), TIMEDIFF(NOW(),s.lastTag) FROM players p left join starve s on p.pId = s.zId; ";
$sql.= $mysqli_real_escape_string("UPDATE standings SET sStatus = "Human" WHERE sId NOT IN (SELECT DISTINCT taggedId FROM tags); "):
$sql.= $mysqli_real_escape_string("UPDATE standings SET sStatus = "Zombie" WHERE (sId IN (SELECT DISTINCT taggedId FROM tags)) OR (sId IN (SELECT DISTINCT taggerId FROM tags)); "):
$que=$mysqli_query($sql);
$sp = mysqli_query($con,"call updateStandings()");
$result = mysqli_query($con,"SELECT * FROM standings");
echo "<table>
<tr>
<th>Id</th>
<th>Name</th>
<th>Status</th>
<th>Since Last Tag</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td style=\"text-align:center\">" . $row['pId'] . "</td>";
echo "<td>" . $row['pName'] . "</td>";
echo "<td style=\"text-align:center\">" . $row['pStatus'] . "</td>";
if ($row['pStarve'] > '72:00:00')
echo "<td style=\"text-align:center\">Starved</td>";
else if
(is_null($row['pStarve']))
echo "<td style=\"text-align:center\">--</td>";
else
echo "<td >" . $row['pStarve'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>

Answer 1

I'm just going to point out where the faults are, which may very well fix your code. If it doesn't, then you will need to read up on the better tutorials.
Regarding this code block:
$sql. = "INSERT INTO standings (`pId`,`pName`,`pStarve`) SELECT p.pId, concat(p.pFirst," ",p.pLast), TIMEDIFF(NOW(),s.lastTag) FROM players p left join starve s on p.pId = s.zId; ";
$sql.= $mysqli_real_escape_string("UPDATE standings SET sStatus = "Human" WHERE sId NOT IN (SELECT DISTINCT taggedId FROM tags); "):
$sql.= $mysqli_real_escape_string("UPDATE standings SET sStatus = "Zombie" WHERE (sId IN (SELECT DISTINCT taggedId FROM tags)) OR (sId IN (SELECT DISTINCT taggerId FROM tags)); "):
$que=$mysqli_query($sql);
The concatenation is incorrect $sql. = those should read as $sql .= which is why you say that the DELETE part works but not the rest following that statement. Had you used or die(mysqli_error($con)) to mysqli_query() it would have signaled the error.
Then you're probably wanting to sanitize input using mysqli_real_escape_string() but that's not how that function works and you don't use a $ sign in front of that function.
It works like this:
$variable = mysqli_real_escape_string($con, $_POST['variable']);
Read up on the function:
http://php.net/manual/en/mysqli.real-escape-string.php
Then you're using $mysqli_query with a dollar sign in front of that function, it's just mysqli_query.
Then you're doing SET sStatus = "Human" that should be single quotes and not double quotes SET sStatus = 'Human' and do the same for the other one.
Having used error reporting and checking for DB error using/adding or die(mysqli_error($con)) to mysqli_query() would have signaled the errors.
Error reporting should have given you warnings such as Undefined variable... in regards to $mysqli_query and $mysqli_real_escape_string.
That should give you enough information to debug your code.

Answer 2

Blank page often means syntax error. Maybe it's the one in front of the INSERT statement?
$sql. = "INSERT INTO standings ...";
Try this one ;)
$sql .= "INSERT INTO standings ...";
Edit: And as mentioned in the comments, it's the first step in debugging. Forgot to say it directly.