PHP計算數據庫中特定條目的行

Question

PHP的新手，被解決類似問題的所有不同解決方案所淹沒。 我不知道我是否有編碼問題，多重查詢問題，或兩者兼有。

在一個php文件中，我打開一個連接，運行一個查詢，然后成功計算該條目出現在數據庫中的次數...或至少嘗試這樣做。

// $team1, $team2 and $page come in through _POST up here...

$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);

// build long query at this point...

$result = mysqli_query($connection, $query);

//I was successful getting it into the database, now I want to count how many times each entry appears.
if ($result) {
        $team1result = mysqli_query($connection,"SELECT * FROM {$page} WHERE 'vote' = {$team1}") ;
        $team1row = mysqli_fetch_row($team1result);
        $team1count = $team1row[0];

        $team2result = mysqli_query($connection,"SELECT * FROM {$page} WHERE 'vote' = {$team2}") ;
        $team2row = mysqli_fetch_row($team2result);
        $team2count = $team2row[0];

        echo $team1count . " and " . $team2count;
}

我可以很好地插入數據庫，但是隨后我的console.log亮起...

警告 ：mysqli_fetch_row（）期望參數1為mysqli_result，布爾值在...中給出
警告 ：mysqli_fetch_row（）期望參數1為mysqli_result，布爾值在...

感謝您今晚的所有幫助。

解決方案（感謝wishchaser）：

if ($result) {
    $team1rows = mysqli_num_rows(mysqli_query($connection,"SELECT * FROM $page WHERE vote = '$team1'"));
    $team2rows = mysqli_num_rows(mysqli_query($connection,"SELECT * FROM $page WHERE vote = '$team2'"));
    echo $team1 . " : " . $team1rows . "   |   ". $team2 . " : ". $team2rows;
}

Answer 1

查詢$ team1result和$ team2result沒有結果行。 這就是為什么您會收到此錯誤。

使用if語句來檢查

if($team1result)
$team1row = mysqli_fetch_row($team1result);

if($team2result)
$team2row = mysqli_fetch_row($team1result);

您將不會得到錯誤。

為了計算查詢結果的行數，請使用以下

$rows=mysqli_num_rows(mysqli_query($query));    

在查詢語句中發現錯誤的一個好習慣是回顯它。

在這種情況下

echo "SELECT * FROM $page WHERE vote = '$team1'";
echo "SELECT * FROM $page WHERE vote = '$team2'";

檢查回顯的查詢是否沒有錯誤（例如未定義的變量）。

Answer 2

您可以使用num_rows輕松計數，無需訪問其索引，只需使用此即可

echo $team1row = mysqli_num_rows($team1result);

參考鏈接