I am getting data from mysql data base using php script problem is that in table there is data but when i fetch returen NO only
it shows following error
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/content/i/h/u/ihus235/html/cs/emrapp/surveyDescription.php on line 73
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HTML CODE
<html<body>
<INPUT TYPE = "Text" VALUE ="user_id" NAME = "user_id"> This form allows you to connect to the server.<br>
<form action="surveyDescription.php" method="post" enctype="multipart/form-data"><br>
Type (or select) Filename:
<input type="submit" value="submit">
</html>
PHP CODE
<?php
$host = "********";
$user = "********";
$pass = "********";
$database = "****";
$linkID = mysql_connect($host, $user, $pass) or die("Could not connect to host.");
mysql_select_db($database, $linkID) or die("Could not find database.");
if (!function_exists('json_encode'))
{
function json_encode($a=false)
{
if (is_null($a)) return 'null';
if ($a === false) return 'false';
if ($a === true) return 'true';
if (is_scalar($a))
{
if (is_float($a))
{
// Always use "." for floats.
return floatval(str_replace(",", ".", strval($a)));
}
if (is_string($a))
{
static $jsonReplaces = array(array("\\", "/", "\n", "\t", "\r", "\b", "\f", '"'), array('\\\\', '\\/', '\\n', '\\t', '\\r', '\\b', '\\f', '\"'));
return '"' . str_replace($jsonReplaces[0], $jsonReplaces[1], $a) . '"';
}
else
return $a;
}
$isList = true;
for ($i = 0, reset($a); $i < count($a); $i++, next($a))
{
if (key($a) !== $i)
{
$isList = false;
break;
}
}
$result = array();
if ($isList)
{
foreach ($a as $v) $result[] = json_encode($v);
return '[' . join(',', $result) . ']';
}
else
{
foreach ($a as $k => $v) $result[] = json_encode($k).':'.json_encode($v);
return '{' . join(',', $result) . '}';
}
}
}
$user_id=$_POST['user_id'];
$query = mysql_query("SELECT s.*, u.user_id FROM survey_master AS s
JOIN user_profile AS u on u.user_id = s.user_id where s.user_id=$user_id");
$rows = array();
while($row = mysql_fetch_assoc($query)) {
$rows[] = $row;
}
echo json_encode($rows);
?>
Problem is in
May be $user_id don’t have any value.please check that.