Print SQL Query Results to Table

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  • kevin11phi
    New Member
    • Oct 2011
    • 7
    #1

    Print SQL Query Results to Table

    Hey Everyone!!

    I am a super noob at php. Also, I am having a bit of a problem with a small project I am working on. The project is simple: use a sql db to list all of the inventory and then test whether or not each machine from the inventory is up (i.e. ping-able). I have only implemented accessing the database but I cannot get php to print the results to a HTML table.

    Here is what I have so far.... :
    Code:
    $con = mysql_connect("localhost", "root", "mysqlr0x");
if (!$con)
{
	die('Could not connect: ' . mysql_error());
}

$query = "SELECT id, name FROM inventory ORDER by basenet_id";
$results = mysql_query($query, $con);

while ($row = mysql_fetch_results($results)) {
	echo "<tr>";
	echo "<td>" .$row['name']."</td>";
	echo "<td>"
	echo "</table>";
}

mysql_close($con);
    Any suggestions?
    Tags: None
    • johny10151981
      Top Contributor
      • Jan 2010
      • 1059
      #2
      I dont know any mysql_fetch_res ults or mysql_fetch_res ult
      Try
      mysql_fetch_row or
      mysql_fetch_ass oc

      there is another
      mysql_fetch_arr ay

        Comment

        • kevin11phi
          New Member
          • Oct 2011
          • 7
          #3
          I updated my code to reflect what you saw... and implemented the ping test (and made it into an html page).
          Code:
          <html>
<head></head>
<body class="page_bg">
<?php #!/usr/bin/php
$con = my_sql_connect("localhost", "root", "dud351");
if (!$con) {
	die('Could not connect: ' . mysql_error();
}
$query = "SELECT id, name FROM inventory ORDER by basenet_id";
$results = mysql_query($query, $con);

function ping($host){
	exec(sprintf('ping -c 1 -W 5 %s', escapeshellarg($host)), $res, 
$rval));
	return rval === 0;
}

while($row = mysql_fetch_assoc($results)) {
	echo "<tr>";
	$checkThisVariable = ping($row['name']);
	if ($checkThisVariable == 1) {
		echo "<td>" . $row['name'] . " is up!</td>";
	}
	else {
		echo "<td>" . $row['name'] . " is down!</td>";
	}
	echo "</table>";
}

mysql_close($con);
?>
</body>
</html>
          Anything else that I am doing wrong?
          However, the code still does not print out a table with the results of my query.

          EDIT: After attempting to run the page from my webserver, I got this:
          Code:
          "; $checkThisVariable = ping($row['name']); if ($checkThisVariable == 1) { echo "" . $row['name'] . " is up!"; } else { echo "" . $row['name'] . " is down!"; } echo ""; } mysql_close($con); ?>
          Last edited by kevin11phi; Oct 25 '11, 11:18 PM. Reason: New stuff

            Comment

            • johny10151981
              Top Contributor
              • Jan 2010
              • 1059
              #4
              your code is not suppose to compile at all, look at line 7

              there is a syntax error.

                Comment

                • kevin11phi
                  New Member
                  • Oct 2011
                  • 7
                  #5
                  I caught the syntax error but I forgot to change it in the code section of my post. The code seems to quit at the point where I need it to print my results associative array. Did I do something wrong?

                    Comment

                    • bektorkhan
                      New Member
                      • Oct 2011
                      • 1
                      #6
                      echo "<td>" . $row['name'] . " is up!</td>";

                      At a quick glance should be:

                      echo "<td>" . $row["name"] . " is up!</td>";

                        Comment

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