Not able to search!?!?!

[nca]

$a1 = "SELECT PHONE, NAME FROM table WHERE NAME LIKE %$name%";
$a2 = mysql_query ("$a1");
$a3 = @mysql_fetch_array ($a2);
if (!$a3) {echo "NOT FOUND";}
echo "$a3[0]<br>";
echo "$a3[1]<br>";

This code will not work. Can anyone see a problem with it?

[RedDragon]

hmm why are you using $a3 = @mysql_fetch_array ($a2);

try with
$a3 = mysql_fetch_array($a2);

also you should use single quotes in your query:

$a1 = "SELECT phone, name FROM table WHERE (name LIKE '%$name%')";

[nca]

The @ hides the ugly mysql error message, if it does not find a match.

I'll give your version with the single quotes a try...

[nca]

$a1 = "SELECT PHONE, NAME FROM table WHERE (NAME LIKE '%$name%')";
$a2 = mysql_query ("$a1");
$NUMROWS = @mysql_num_rows($a2);
if (!$NUMROWS) {$notfound=1;}
if ($NUMROWS) {
$I = 0;
while ($I < $NUMROWS) {
$a3 = @mysql_fetch_array ($a2);
echo "$a3[0] $a3[1]<br>";
}
}

Why doesn't this work????? When you put single quotes in a variable doesn't the $a2 see it as '%$name%' and not with the variable name?

[altexis]

try this one:

$a1 = "SELECT PHONE, NAME FROM table WHERE NAME LIKE '%".mysql_escape_string($name)."%'";
$a2 = mysql_query ("$a1");
if ($a3 = mysql_fetch_array ($a2)) {
  echo "<p>";
  echo "$a3[0]<br>";
  echo "$a3[1]<br>";
  echo "</p>\n";
}
else echo "NOT FOUND";

[nca]

It only shows the first one found. 🙁
I need it to show all matches.

[GSHelpBoy]

See if this works:

$a1 = "SELECT `phone`, `name` FROM `table` WHERE `name` LIKE '%" . $name . "%'";
$a2 = mysql_query ($a1);
if ($a3 = mysql_fetch_array ($a2)) {
  echo "<p>";
  echo "$a3[0]<br>";
  echo "$a3[1]<br>";
  echo "</p>\n";
} else {
  echo "NOT FOUND";
}

It's minor changes like using backquotes (`) for the field names and the table.

[altexis]

oook... then try this one 🙂

$a1 = "SELECT PHONE, NAME FROM table WHERE NAME LIKE '%".mysql_escape_string($name)."%'";
$a2 = mysql_query ("$a1");
if (mysql_num_rows($a2)==0) echo "NOT FOUND";
else while ($a3 = mysql_fetch_array ($a2)) {
echo "<p>";
echo "$a3[0]<br>";
echo "$a3[1]<br>";
echo "</p>\n";
}

[nca]

THANK YOU!!! 😃