Question

我希望我的页面只加载一次，发生的事情是它不断重新加载，请帮我解决这个问题。

我希望代码在页面加载时立即运行，但我不希望它重新加载，它会继续重新加载我的页面。

除了重新加载问题以外，一切都很完美。

<!DOCTYPE html>

<html>

<head>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>

</head>

<body onload="getLocation()">

<script>

var x = document.getElementById("demo");

function getLocation() {
    if (navigator.geolocation) {
        navigator.geolocation.getCurrentPosition(redirectToPosition);
    } else { 
        x.innerHTML = "Geolocation is not supported by this browser.";
    }
}

function redirectToPosition(position) {
    window.location='j3.php?lat='+position.coords.latitude+'&long='+position.coords.longitude;
}
</script>

<?php

$lat=(isset($_GET['lat']))?$_GET['lat']:'';
$long=(isset($_GET['long']))?$_GET['long']:'';

$url = "https://maps.googleapis.com/maps/api/geocode/json?latlng=" . $lat . "," . $long;

function get_data($url) {
    $ch = curl_init();
    $timeout = 5;
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, $timeout);
    $data = curl_exec($ch);
    curl_close($ch);
    return $data;
}

$returned_contents = get_data($url);

$returned_contents_array = json_decode($returned_contents, true);
$results = $returned_contents_array['results'];

foreach($results as $result) {

   echo $result['address_components'][0]['long_name'] . ", <br>";
}

?>

</body>
</html>

Answer 1

如何检查坐标是否在函数内设置：

function getLocation() {
    <?php if (!isset($_GET['lat']) || !isset($_GET['long'])) { ?>
    if (navigator.geolocation) {
        navigator.geolocation.getCurrentPosition(redirectToPosition);
    } else { 
        x.innerHTML = "Geolocation is not supported by this browser.";
    }
    <?php } ?>
}

这样，如果两个坐标参数都已设置，则函数体将为空并且不执行任何操作。

我希望页面只加载一次

1 个答案: