Php o我需要推送到一个数组，然后将其作为输出进行循环？谢谢。。那么我可以在一个循环中“构建”它吗？ad是否让它在找到/匹配数字的每个记录中返回一个值？感谢这样：从har_id所在的位置选择“guar_num”（“按数字列表构建……有1000个”）从har_id

---

o我需要推送到一个数组，然后将其作为输出进行循环？谢谢。。那么我可以在一个循环中“构建”它吗？ad是否让它在找到/匹配数字的每个记录中返回一个值？感谢这样：

从har_id所在的位置选择“guar_num”（“按数字列表构建……有1000个”）

从har_id所在的位置选择“guar_num”（“按数字列表构建……有1000个”）

这是我目前的位置：`harNumArray2=内爆（'，'，$harNumArray）；//$harNumArray2=''”。内爆（“，”，$harNumArray2）。“；$query=“从har_id所在的位置选择guar_nu（$harNumArray2）”；echo$query；'这就是我现在的位置：

harNumArray 2=内爆（'，'，$harNumArray）；//$harNumArray2=''”。内爆（“，”，$harNumArray2）。“；$query=“从har_id所在的位置选择guar_nu（$harNumArray2）”；echo$query；

$harNumArray2 = implode(',', $harNumArray);
//$harNumArray2 = '"' . implode('","', $harNumArray) . '"';
$query = "SELECT guar_nu FROM placements WHERE har_id IN ($harNumArray2)";
echo $query;

$match = mysql_query($query);
//$match = mysql_query('"' . $query . '"');
$results = $match;
echo("<BR><BR>");
print_r($results);

$harNumArray2 = implode(',', $harNumArray);
//$harNumArray2 = '"' . implode('","', $harNumArray) . '"';
//$query = "'SELECT guar_num FROM placements WHERE har_id IN ($harNumArray2)'";

$result = mysql_query("SELECT har_id, guar_num FROM placements WHERE har_id IN (" . $harNumArray2 . ")")
//$result = mysql_query("SELECT har_id, guar_num FROM placements WHERE har_id IN (0108383442,0106620416)")
or die(mysql_error());  

// store the record of the "example" table into $row
$row = mysql_fetch_array($result);
$numRows = mysql_num_rows($result);

/*
while($row = @mysql_fetch_assoc($result) ){
    // do something
    echo("something <BR>");
}
*/

// Print out the contents of the entry 
echo("TOTAL ROWS RETURNED : " . $numRows . "<BR>");
echo "HAR ID: ".$row['har_id'];
echo " GUAR ID: ".$row['guar_num'];

SELECT * FROM tbl1 WHERE id = (SELECT num FROM tbl2 WHERE id = 1);

SELECT * 
FROM YOUR_TABLE 
WHERE COLUMN_NAME IN (LIST, OF, SEARCH, VALUES, SEPARATED, BY COMMAS)

  $mysqli   = mysqli_connect("localhost", "user", "pass");
  mysqli_select_db($mysqli, "YOUR_DB");

  // make a comma separated list of the $ids.
  $ids = join(", ", $id_list);

  // note: you need to pass the db connection to many of these methods with the mysqli_ API
  $results  = mysqli_query($mysqli, "SELECT har_id, guar_num FROM placements WHERE har_id IN ($ids)");
  $num_rows = mysqli_num_rows($results);

  while ($row = mysqli_fetch_assoc($results)) {
    echo "HAR_ID: ". $row["har_id"]. "\tGUAR_NUM: " . $row["guar_num"] . "\n";
  }