Question

我正在尝试从mysql数据库中选择一个SINGLE值。我在phpmyadmin中运行了查询，效果很好。但是当我回应$ result时，我什么都没得到......顺便说一下，对于我使用xxx的数据库和密码，因为我不想显示它...我的插入查询工作得很好

由于

<?php

//Create Connection
$servername = "localhost";
$username = "root";
$password = "xxx";
$dbname = "xxx";



//Connect
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT StartPriceUnder FROM YJ_Value";
$result = $conn->query($sql);



echo hi;
echo $result;
echo ya;

$conn->close();

?>

Answer 1

试试这个：

<?php
$servername = "localhost";
$username = "root";
$password = "xxx";
$dbname = "xxx";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT StartPriceUnder FROM YJ_Value";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
    while($row = $result->fetch_assoc()) {
        echo "StartPriceUnder:" . $row["StartPriceUnder"];
    }
} 
else {
    echo "0 results";
}
    $conn->close();
?>

Answer 2

您必须获取结果，所以请执行以下操作：

$row = $result->fetch_array(MYSQLI_ASSOC);

在此之后你可以像这样回应它：

echo $row["StartPriceUnder"];

有关fetch_array()的详情，请参阅手册：http://php.net/manual/en/mysqli-result.fetch-array.php

php无法选择数据

2 个答案: