Question

mysql_fetch_array无效。一切都很好看。我不知道我在做错什么。

$sql = "SELECT * FROM `$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
 {
    }
}

有人可以帮助我在mysql_fetch_array找到我的错误吗？

Answer 1

您应该尝试非常仔细地理解以下代码。这个目前正在我的社交网站上使用，我希望这能解决你的问题。

$st= "SELECT* FROM users WHERE username='$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 2

尝试这种方式......我认为你正在搞乱单引号和双引号

$sql = "SELECT * FROM `$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
 }
}

Answer 3

您可以查看查询中是否有任何错误：mysql_error(); 此函数返回与mysql相关的函数中的最后一个错误。只需使用以下语法：

$resultw = mysql_query($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
    ...
}

执行mysql_fetch_array时出错

3 个答案: