无法从 PHP 访问数据库值

Question

我已经从 php 连接到 mySql 数据库。我想从中获取两列的详细信息。当我尝试访问它们时，它输出 null。当我在 phpmyAdmin 数据库工具中运行时，相同的查询字符串工作正常。

下面是我的代码：

$servername = "mysql";
            $username = "$$$$$$$$$";
            $password = "%%%%%%%%%%";
            $databasename = 'blog';

            // Create connection
            $conn = new mysqli($servername, $username, $password, $databasename);

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



 $select_sql = "Select site_url from sites where id LIKE '%website1Modal%'";
            $result = $conn->query($select_sql);

            if ($result->num_rows > 0) {
              echo "in if loop";
               echo $result[0];
}

以上代码输出："Connected successfully" 和 "in if loop".

有人能告诉我我犯了什么愚蠢的错误吗？

Answer 1

由于 $result[0] 是一个数组，您不能使用 echo 显示它，而是使用 print_r :

echo '<pre>';
print_r($result[0]);
echo '</pre>';

如果你想显示 site_url var 你需要使用 :

echo $result[0]['site_url'];

Answer 2

if 不会让您陷入循环。 if 是一个控制结构。使用 while 进入循环并将 fetch 放入其中，以便可以访问结果。

试一试：

$servername = "mysql";
$username = "$$$$$$$$$";
$password = "%%%%%%%%%%";
$databasename = 'blog';
// Create connection
$conn = new mysqli($servername, $username, $password, $databasename);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} else {
    echo "Connected successfully";
}
$select_sql = "Select site_url from sites where id LIKE '%website1Modal%'";
$result = $conn->query($select_sql);
if ($result->num_rows > 0) {
    echo "in if control";
    while($row = $result->fetch_array()) { //this is optional; if only one row the while can be removed. e.g. $row = $result->fetch_array();
        echo 'in while loop';
        echo $row['site_url'];
    } //if removing while remove this as well
}

您可以在此处阅读有关 if 和 while 控件的更多信息。

http://php.net/manual/en/control-structures.if.php
http://php.net/manual/en/control-structures.while.php

这是 query 功能的手动条目。

重要的一点

For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object.

并且因为您有 mysqli_result object，所以您需要将其发送到 fetch，后者将

Returns an array of strings that corresponds to the fetched row or NULL if there are no more rows in resultset.

Answer 3

您需要使用 Mysqli_Result::fetch_assoc()

获取结果的每一行

while($row = $result->fetch_assoc())
{
    print_r($row);
}

将打印结果集的每一行。

您可以访问循环内每个属性结果行；

$row['col_name'];

其中 col_name 是列的名称。

Answer 4

// put your db-connection in a different file, preferably placing it outside your webroot
include('../some/folder/dbconn.php');

$select_sql = "Select site_url from sites where id LIKE '%website1Modal%'";
$result = $conn->query($select_sql);

if ($result->num_rows() > 0) {
    // fetch the result
    $result = $conn->fetch();
    // echo the result
    echo $result['site_url'];
}

// close db connection
$conn = NULL;

无法从 PHP 访问数据库值

Cannot acces the database value from PHP

php

mysql

resultset