Question

我要求我的数据库提供数据,并把它放入阵列,因为我想在路外使用。问题在我的第二个问询中是问问题。

$cari2 = "SELECT name, data FROM `dns_rr` WHERE `zone` =  ".$row[ id ]."  LIMIT 0, 30 ";
$keputusan2 = mysql_query($cari2);

while($row2 = mysql_fetch_array($keputusan2)) {
//echo $row2[ name ];
$arrayData[] = $row2[ data ];
$arrayName[] = $row2[ name ];
}

Array output:
Array ( [0] => gumblari.cn [1] => gumblar.cn ) Array ( [0] => gumblaro.cn [1] => okla )

I also use implode before running the query:
$implodeName = implode(", ",$arrayName); $implodeData = implode(", ",$arrayData);

Implode output:
gumblaro.cnoklagumblari.cngumblar.cn

   $cari3 = "SELECT data, name FROM `dns_rr` WHERE data IN ($implodeData,      $implodeName)";
$keputusan3 = mysql_query($cari3);

while($row3 = mysql_fetch_array($keputusan3)) {

echo $row3[ data ];
echo $row3[ name ];

}

Query no 2 ($cari3) give me this error:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in

Help me please. What I need is to use the data from first query outside while loop thats why I resorted to array. Thank you.

Answer 1

忽视其他贵方的文字问题(如卡片注射),由于盘问失败, $子2美元的价值很可能是一种ool。你们需要确保3美元有效。我的猜测是,这两个变量都是引出的,被解读为是Q8而不是价值。你们需要确保这两个阵列都像价值1,价值2。您可以使用“......,......,“,......,$array”,“;基本上,在进行询问之前,先印刷,看错。另见我的sql_error()所说的话。

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