never ending while loop

[Foser]

<?php
mysql_connect("localhost", "root", "password") or die(mysql_error);
mysql_select_db("tutorial1") or die(mysql_error);


echo "<table border='1'>";
echo "<tr><th>Name</th><th>Age</th><th>Sex</th></tr>";

while($personalinfo = mysql_fetch_array(mysql_query("SELECT * FROM personalinfo")) or die(mysql_error())){

echo "<tr><td>";
echo $personalinfo['Name'];
echo "</td><td>";
echo $personalinfo['Age'];
echo "</td><td>"; 
echo $personalinfo['Sex'];
echo "</td></tr>"; 
}
echo "</table>";
?>

 

im trying to make it say it once, and if there another different one that it goes through that and if there is no more after that, to stop. But it just goes on, until a 30 sec error.

 

[clown[NOR]]

maybe something like this

 

<?php

$result = mysql_query("SELECT * FROM personalinfo");
if (!$result) { die(mysql_error()); }

$dbRows = mysql_num_rows($result);

if ($dbRows > 0) {

echo "<table border='1'>";
echo "<tr><th>Name</th><th>Age</th><th>Sex</th></tr>";

while ($perinfo = mysql_fetch_assoc($result)) {

echo "<tr><td>";
echo $personalinfo['Name'];
echo "</td><td>";
echo $personalinfo['Age'];
echo "</td><td>"; 
echo $personalinfo['Sex'];
echo "</td></tr>"; 
}
echo "</table>";
} else {
echo "Sorry. No info found.";
}
?>

[Foser]

i've read a few article about mysql_num_rows() but I can't find exacly in stupid meanings what it does exacly. can anyone help an idiot? 

 

also on this line

if (!$result) { die(mysql_error()); }

  why the exaclamation point?

[per1os]

mysql_num_rows counts the number of rows the query uses to produce the query

 

So technically if no rows are returned you have an empty result which means there is no need to do the while stuff.

 

The reason your solution did not work was because the mysql_query() was re-generating the query in your while loop, so it would only go to the first record. Where as clown's solution the mysql_query is assigned to the variable and is not re-generated each time the loop is ran.

 

Hope that helps.

[Foser]

mysql_connect("localhost", "root", "password") or die(mysql_error);
mysql_select_db("tutorial1") or die(mysql_error);

$sqlquery = mysql_query("SELECT * FROM personalinfo");

if (!$sqlquery) { die(mysql_error());}

$dbrows = mysql_num_rows($result);
if ($dbrows > 0) { 
echo "<table border='1'>";
echo "<tr><th>Name</th><th>Age</th><th>Sex</th></tr>";
}
else { 
echo "Data is not available at this time. Please try again later.";

while($personalinfo = mysql_fetch_assoc($sqlquery)){

echo "<tr><td>";
echo $personalinfo['Name'];
echo "</td><td>";
echo $personalinfo['Age'];
echo "</td><td>"; 
echo $personalinfo['Sex'];
echo "</td></tr>"; 
echo "</table>";
}

 

 

got a $end error which is normally a punctuation error.

[papaface]

else { 
echo "Data is not available at this time. Please try again later.";

while($personalinfo = mysql_fetch_assoc($sqlquery)){

Theres no ending }

Should be:

else { 
echo "Data is not available at this time. Please try again later.";
}
while($personalinfo = mysql_fetch_assoc($sqlquery)){

[Foser]

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in

 

 

thats the error im geting

[papaface]

These are simple errors:

Should be:

$sqlquery = mysql_query("SELECT * FROM personalinfo");

if (!$sqlquery) { die(mysql_error());}

$dbrows = mysql_num_rows($sqlquery);

[Foser]

yea that was embarrassing im a tad sleepy.

 

last question

 

for some reason my first is in the table, but the second is not. its outside of the table...

[Foser]

bump

[Caesar]

bump

yea that was embarrassing im a tad sleepy.

 

last question

 

for some reason my first is in the table, but the second is not. its outside of the table...

 

<?php

while($personalinfo = mysql_fetch_assoc($sqlquery)){

echo "<tr><td>";
echo $personalinfo['Name'];
echo "</td><td>";
echo $personalinfo['Age'];
echo "</td><td>"; 
echo $personalinfo['Sex'];
echo "</td></tr>"; 
echo "</table>"; //<---- THIS IS WHY
}

?>