Question

我想通过以下代码显示我周围最近的位置

<?php
    $lat = "27.7350758";
    $long = "85.3102946";

    $result = $conn->query( 'SELECT * , (3956 * 2 * ASIN(SQRT( POWER(SIN(( $lat - college_lat) * pi()/180 / 2), 2) +COS( $lat * pi()/180) * COS(college_lat * pi()/180) * POWER(SIN(( $long - college_lon) * pi()/180 / 2), 2) ))) as distance from colleges having distance <= 10 order by distance' );

                if ( $result->num_rows>0 ) {
                    // output data of each row
                    while( $row = $result->fetch_assoc() ) {
                        echo '<div class="category"><div class="cover" style="background-image: url(img/animation.jpg);"><div>'; ?>    
                        <td><?php echo $row[ "college_name" ]; ?></td>  
                        <?php
                        echo '</div></div><span class="counter ">Courses : <span>' . $row[ "college_courses" ] . '</span></span>';
                        echo '<span class="counter "> Distance : <span>' . $row[ "college_address" ] . '</span></span>';
                        echo '<span class="near-you"> <p>'. substr( $row[ "college_desc" ], 0 , 100) . '</p> </span></div>'; ?>
                        <a href="single.php?college_id=<?php echo $row['college_id'];?>"> Read More... </a>

                <?php      
                    }
                } else {
                    echo "Colleges Not Found Please add";
                }
              ?>

当我运行此代码时，将显示以下错误

注意：尝试在第56行的C：\ xampp \ htdocs \ _city \ index.php中获取非对象的属性未找到大学请添加

如果我删除$ lat和$ long，那么保持查询值可以正常工作。我的错误是什么，请建议我。谢谢

Answer 1

更改您的查询，如下所示。将'更改为"，否则不会解释值：

$result = $conn->query( "SELECT * , (3956 * 2 * ASIN(SQRT( POWER(SIN(( $lat - college_lat) * pi()/180 / 2), 2) +COS( $lat * pi()/180) * COS(college_lat * pi()/180) * POWER(SIN(( $long - college_lon) * pi()/180 / 2), 2) ))) as distance from colleges having distance <= 10 order by distance" );

What is the difference between single-quoted and double-quoted strings in PHP?

Answer 2

我认为，这是因为，值$lat和$long被视为字符串。您可以更好地使用预准备语句并绑定这些变量。

$stmt = $mysqli->prepare("SELECT * , (3956 * 2 * ASIN(SQRT(POWER(SIN((? - college_lat) * pi()/180 / 2), 2) + COS(? * pi()/180) * COS(college_lat * pi()/180) * POWER(SIN(( ? - college_lon) * pi()/180 / 2), 2) ))) as distance from colleges having distance <= 10 order by distance");
$stmt->bind_param('ddd', $lat, $lat, $long);

获取错误尝试获取非对象的属性

2 个答案: