執行mysql_fetch_array時出錯

Question

mysql_fetch_array不起作用。一切都很美好。我不知道我在哪裏做錯了什麼。

$sql = "SELECT * FROM `$tbl_name` limit $start,$limit";//if echo gives o/p Resource id #14 

$resultw = mysql_query($sql); 
while($gup=mysql_fetch_array($resultw))//if echo gives o/p Array 
{ 
//if echo $gup['to']; gives o/p vicky.0989hyd@gmail.com; 

$anusha=mysql_query("select * from users where email='$gup[to]'");//if echo gives o/p Resource id #15 

    while($resulter = mysql_fetch_array('$anusha'))//here is what iam getting error Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in 
{ 
    } 
} 

有人能幫我找出我的錯誤在我的mysql_fetch_array？

Answer 1

搞亂你可以檢查，看看是否有您所查詢的任何錯誤與：mysql_error(); 這個函數返回mysql相關函數中的最後一個可能的錯誤。 只要使用這個語法：

$resultw = mysql_query($sql) or die(mysql_error()); 
$gup=mysql_fetch_array($resultw) or die(mysql_error()); 
while($gup)//if echo gives o/p Array 
{ 
    $email=$gup['to']; 
    $anusha=mysql_query("select * from users where email='$email'") or die(mysql_error()); 
    //check $anusha here before pass it to mysql_fetch_array function 
    ... 
} 

Answer 2

您應該非常仔細地嘗試瞭解以下代碼。這是目前我的社交網站上使用的，我希望這可以解決您的問題。

---

$st= "SELECT* FROM users WHERE username='$you'"; 
$result=mysqli_query ($con,$st); 
$count=mysqli_num_rows ($result); 

if($count==0){echo "<b>Profile not found! </b> ";} 
else {echo "<b>Your Profile..</b>";} 
while($row= mysqli_fetch_array ($result)) { echo "Username-<b>". $row ['uname']. "</b>"; echo "sent you this message"; 
echo "<p id='sms'>". $row ['sms']. "</p>"; 

Answer 3

試試這個方法...我認爲你是有單，雙引號

$sql = "SELECT * FROM `$tbl_name` limit $start,$limit"; 
$resultw = mysql_query($sql); 
if (!$resultw) { 
die('Invalid query: ' . mysql_error()); 
} 
while($gup=mysql_fetch_array($resultw))//if echo gives o/p Array 
{ 
$email=$gup['to']; 
$anusha=mysql_query("select * from users where email='$email'"); 
if (!$anusha) { 
die('Invalid query: ' . mysql_error()); 
} 
//check $anusha here before pass it to mysql_fetch_array function 
while($resulter = mysql_fetch_array($anusha)) 
{ 
    //do what you want to do man 
} 
}