Question

早上好，我对此问题进行了广泛的研究，但是不幸的是，没有任何相关问题解决了我的问题。

这里，我有一个非常基本的PHP mySQLi db连接。连接成功，在表上运行的查询也将成功。问题是结果集将不会显示。我所有的引用都是正确的，当我检查是否填充了结果集时，它就是正确的。我相信问题出在我的while块上，但是运行时不会返回任何错误。谢谢您的时间

<?php
$db = mysqli_connect('localhost','root','','securitour') //connection to the         database
or die('Error connecting to MySQL server.');
?>

<html>
<head>
</head>
<body>
<?php

$query = "SELECT * FROM location"; //The SQL query
mysqli_query($db, $query) or die('Error querying database.'); 
$result = mysqli_query($db, $query); //query the table an store the result set in a     variable
$row = mysqli_fetch_array($result); //create an array and store the records of the     result set in it

if (mysqli_num_rows($result) != 0) //to check if the result set contains data
{
    echo "results found"; //THIS is what is returned.
} 
else 
{
    echo "results not found";
}
while ($row = $result->fetch_assoc()) //itterate through the array and display the         name column of each record
    {
    echo $row['name'];
    }
    mysqli_close($db);
?>
</body>
</html>

Answer 1

您无需运行mysqli_query()两次。并且您需要将mysqli_fetch_assoc用于关联数组

<?php
$db = mysqli_connect('localhost','root','','securitour') or die('Error connecting to MySQL server.');
?>

<html>
<head>
</head>
<body>
<?php
$query = "SELECT * FROM location"; //The SQL query
$result = mysqli_query($db, $query) or die('Error querying database.'); //query the table an store the result set in a     variable
$row = mysqli_fetch_assoc($result); //create an array and store the records of the     result set in it

if (mysqli_num_rows($result) != 0) //to check if the result set contains data
{
    echo "results found"; //THIS is what is returned.
} else {
    echo "results not found";
}

foreach ( $row as $name=>$val) {
    echo $name . ':' . $val . '<br>';
}
mysqli_close($db);
?>
</body>
</html>

Answer 2

很多东西不在这里。

您正在处理mysqli_query（）函数两次-不需要。
您正在选择SQL查询（SELECT *）中的所有字段。您应该按名称选择字段。
您正在过程和基于类的MySQLi之间互换-您应该坚持一个或另一个。

尝试以下方法：

    <?php
    $db = mysqli_connect('localhost','root','','securitour') //connection to the         database
    or die('Error connecting to MySQL server.');
    ?>

    <html>
    <head>
    </head>
    <body>
    <?php

    $query = "SELECT name FROM location"; //The SQL query
    $result = mysqli_query($db, $query) or die('Error querying database'); //query the table an store the result set in a     variable
    if(mysqli_num_rows($result) > 0){
        echo "Results found!";
        while($row = mysqli_fetch_array($result)){ //create an array and store the records of the     result set in it
            echo $row['name'];
        }
    } else {
        echo "results not found";
    }
        mysqli_close($db);
    ?>
    </body>
    </html>

PHP mySQLi查询返回数据但不显示数据

2 个答案: