Question

我正在尝试检查leaguename名称是否已存在，如果确实存在，则检查它是否高于现有分数。如果它更高，则替换quizscore否则将leaguename和quizscore插入表league_quiz。

代码似乎在我的表中插入了值，但是如果名称等于previous并且分数更高，它似乎不会更新？

我收到此错误：

警告：mysqli_num_rows()期望参数1为mysqli_result，字符串在
中给出

我的代码：

<?php

$name = (string)$_POST['name'];
$score = (string) $_POST['score'];

$link = mysqli_connect("mysql12.gigahost.dk","username","password","dirts_mysql");

$query = "SELECT leaguename, quizscore FROM league_quiz WHERE  leaguename = '$name' AND quizscore < '$score'";

if(mysqli_num_rows($query)) {
    mysqli_query($link,"UPDATE league_quiz SET quizscore = '$score'");
} else {
    mysqli_query($link,"INSERT INTO league_quiz (leaguename, quizscore) VALUES ('$name', '$score')") or die(mysqli_error($link));
}

?>

Answer 1

在这两者之间：

$query = "SELECT leaguename, quizscore FROM league_quiz WHERE  leaguename = '$name' AND      quizscore < '$score'";

if(mysqli_num_rows($query)) {

您是否尝试过编写此代码？

$result = mysqli_query($query);

然后对查询结果使用mysqli_num_rows？

Answer 2

mysqli_num_rows需要查询结果，而不是查询本身的字符串。首先，您必须执行查询：

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

然后你可以计算行数

$numRows = mysqli_num_rows($result);

Answer 3

尝试以下方法：

$result = mysqli_query($query);
if (mysqli_num_rows($result)) {
...

更新并插入mysqli表

3 个答案: