Question

我有这个查询在同一个网站的其他部分工作正常，但我遇到的问题是检查说不是真的，但它实际上是否正确执行？

<?php

    $conn = new mysqli($mysql_host, $mysql_user, $mysql_password, $mysql_database);
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 

    $sql = "SELECT * FROM Stock WHERE StockItemID='12'";
    $result = $conn->query($sql) or die( $conn->connect_error );
    $row = $result->fetch_assoc();

    if ($conn->query($sql) === TRUE) {
         echo $row['StockCategory'];
    } else {
         printf("Errorcode: %d\n", $conn->errno);
    }
    $conn->close();
?>

因为在上面的例子中我没有得到回声，我认为它没有用但我回来了 - 错误代码：0

但是，如果我从表中回显出一个值，它就会起作用。

Answer 1

选择查询返回结果集

$conn = new mysqli($mysql_host, $mysql_user, $mysql_password, $mysql_database);
if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } 
$sql = "SELECT * FROM Stock WHERE StockItemID='12'";
$result = $conn->query($sql) or die( $conn->connect_error );
if ($result->num_rows > 0) {
    $row = $result->fetch_assoc();
    echo $row['StockCategory'];
} else {
    printf("Errorcode: %d\n", $conn->errno);
}
$conn->close();

http://php.net/manual/en/mysqli.query.php

Answer 2

如果您想查看查询是否返回一行，那么检查该行本身。

是的，尽可能简单和合乎逻辑。

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = new mysqli($mysql_host, $mysql_user, $mysql_password, $mysql_database);
$sql = "SELECT * FROM Stock WHERE StockItemID='12'";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
if ($row) { // here
    echo $row['StockCategory']; 
}

查询false但执行查询

2 个答案: