PHP - 代码几乎达到了我想要的效果

Question

下面的代码呼应了以下内容:

Table Name 1

Table Name 2

Table Name 3

Table Name 4 "$entry": "votes_up for $entry in Table Name 4"

我希望它能回应这个:

Table Name 1: "votes_up for $entry in Table Name 1"

Table Name 2: "votes_up for $entry in Table Name 2"

Table Name 3: "votes_up for $entry in Table Name 3"

Table Name 4: "votes_up for $entry in Table Name 4"

如何更改代码以使其符合我的要求？

表名称的数量根据 $entry 的不同而变化。

谢谢

约翰

$result = mysql_query("SHOW TABLES FROM feather") 
or die(mysql_error()); 

while(list($table)= mysql_fetch_row($result))
{
 $sqlA = "SELECT `site`,votes_up FROM `$table` WHERE `site` LIKE '$entry'";
 $resA = mysql_query($sqlA) or die("$sqlA:".mysql_error());
 if(mysql_num_rows($resA) > 0)
 {
$table_list[] = $table;
while($rowA = mysql_fetch_assoc($resA))
  {
  $votes_up[$rowA["site"]] = $rowA["votes_up"];
  }
 }
}

foreach( $table_list as $key => $value){
    echo "$value <br />";
}

foreach($votes_up as $site => $vote_up)
{
  echo "$site: $vote_up";
}

Answer 1

没关系。

这给了我我想要的:

$result = mysql_query("SHOW TABLES FROM feather") 
or die(mysql_error()); 

while(list($table)= mysql_fetch_row($result))
{
  $sqlA = "SELECT COUNT(*) FROM `$table` WHERE `site` LIKE '$entry'";
  $resA = mysql_query($sqlA) or die("$sqlA:".mysql_error());
  list($isThere) = mysql_fetch_row($resA);
  if ($isThere)
  {
     $table_list[] = $table;
  }
}


foreach ($table_list as $table) { 
    $sql = "SELECT votes_up FROM `$table` WHERE `site` LIKE '$entry'"; 
    $sql1 = mysql_query($sql) or die("$sql:".mysql_error());
   while ($row = mysql_fetch_assoc($sql1)) {
       echo $table . ': "' . $row['votes_up'] . " for $entry from $table\"<br />";
   } 
}