这是我在DB_Functions.php中的函数,我想从单个函数中的两个不同的表中获取两个不同的值,这里是我到目前为止尝试过的代码,但是值为空。
public function getUserMetvalue($exname,$fname) {
$result = mysql_query("SELECT metvalue FROM fitnessactivitylist WHERE activityname='$exname'") or die(mysql_error());
$result1 = mysql_query("SELECT weight FROM users WHERE name='$fname'") or die(mysql_error());
// check for result
$no_of_rows = mysql_num_rows($result);
$no_of_rowss = mysql_num_rows($result1);
if ($no_of_rows > 0) {
$result = mysql_fetch_array($result);
if ($no_of_rowss > 0) {
$result1 = mysql_fetch_array($result1);
return $result1;
}
return $result;
} else {
//exercise name not found
return false;
}
}
这是我的index.php
//TAG METVALUE
if ($tag == 'metvalue') {
$exname = $_POST['exname'];
$fname = $_POST['fname'];
$usermetvalue = $db->getUserMetvalue($exname,$fname);
if ($usermetvalue != false) {
$response["success"] = 1;
$response["usermetvalue"]["exname"] = $usermetvalue["exname"];
$response["usermetvalue"]["fname"] = $usermetvalue["fname"];
echo json_encode($response);
}
else {
$response["error"] = 1;
$response["error_msg"] = "No exercise found!";
echo json_encode($response);
}
}
答案 0 :(得分:1)
发现差异:
$result = mysql_query("SELECT metvalue etc...
^^^^^^^^
$result1 = mysql_query("SELECT weight etc...
^^^^^^
$response["usermetvalue"]["exname"] = $usermetvalue["exname"];
^^^^^^
$response["usermetvalue"]["fname"] = $usermetvalue["fname"];
^^^^^
您获取以后未使用的字段,然后尝试访问首先未提取的字段...
答案 1 :(得分:0)
function a_function() {
$a = 'Learn';
$b = 'Programming.';
return array($a, $b);
}
list($one, $two) = a_function();
echo $one . ' ' . $two;