Question

我在将PHP字符串传递给Javascript函数时遇到问题，我读了很多关于这个问题的帖子，并尝试了几种方法，但没有一个对我有用。基本上我有两个函数，一个用PHP编写，从数据库中提取信息，另一个用Javascript编写，旨在让我为google地图编写地址并传递一些信息添加到信息窗口，两位代码是如下所示：

PHP

try {
        $bubbleData = $dbConnection->getBubbleData();
    } catch (Exception $e) {
        echo "The following error occoured while attempting to get Google map info window data" .
        " " . $e->getMessage() .
        " " . "In file" .
        " " . $e->getLine() .
        " " . "on line" .
        " " . $e->getLine();
    }

    if (mysql_num_rows($bubbleData) == 0) {
        echo "No Placement data found to populate map";
    } else {
        while ($row = mysql_fetch_array($bubbleData)) {
            $companyName = $row['Company_Name'];
            $title = $row['Title'];
            $address_1 = $row['Address_Line_1'];
            $address_2 = $row['Address_Line_2'];
            $address_3 = $row['Address_Line_3'];
            $address_4 = $row['Address_Line_4'];
            $post_Code = $row['Post_Code'];
            $forename = $row['Forename'];
            $surname = $row['Surname'];

            $fullAddress = $address_1 . " " . $address_2 . " " . $address_3 . " " . $address_4 . " " . $post_Code;

            $partAddress_1 = $address_1 . " " . $address_2;
            $partAddress_2 = $address_3 . " " . $address_4;

            $infoText = "<h4>" . $title . "</h4>" .
                    '<b>' . "Company name:" . '</b>' . " " . $companyName .
                    "</br>" .
                    "<b>" . "Employee name:" . '</b>' . " " . $forename . " " . $surname . '</br>' .
                    "<b>" . "Company address:" . "</b>" . " " . $partAddress_1 . '</br>' .
                    $partAddress_2 . '</br>' . $post_Code;

    echo "<SCRIPT LANGUAGE='javascript'>
    geocodeAddress(<?php echo json_encode($fullAddress);?>,<?php echo json_encode($infoText);?>);
    </SCRIPT>'";
        }
    }

的Javascript

function geocodeAddress (address,infoText) {

            var geocoder = new google.maps.Geocoder();
            geocoder.geocode( {'address': address}, function(results, status) {
                if (status == google.maps.GeocoderStatus.OK) {
                    loadMarker(results[0].geometry.location,infoText,address);
                    latlngArray.push(results[0].geometry.location);
                } else {
                    alert("Geocode was not successful for the following reason: " +" "+  status);
                }
            });
        }re

现在我知道这两个单独工作正常，但每当我尝试在PHP旁边调用javascript并传递变量时我在firebug中出现以下错误：

属性地址的值无效：

我一直试图让这一切都发生，我使用了正则表达式的平面回声和json_encode（）函数，并且没有人能帮助吗？

提前致谢

Answer 1

您在PHP代码中嵌套了<?php ?>，这不会被解释为PHP。相反，它只是作为字符串打印到您的输出中。

 // Instead of
 echo "<SCRIPT LANGUAGE='javascript'>
    geocodeAddress(<?php echo json_encode($fullAddress);?>,<?php echo json_encode($infoText);?>);
    </SCRIPT>'";

 // Change to
 echo "<SCRIPT LANGUAGE='javascript'>
    geocodeAddress(" . json_encode($fullAddress) . "," . json_encode($infoText) . ");
    </SCRIPT>'";

Answer 2

至少在我的情况下，我注意到了

<?php ...foo... ?>

HTML主体中的

段，仅当文件扩展名为.php时才进行php评估。也就是说，我不得不重命名：

mv foo.html foo.php

将php值传递给javascript变量。

将PHP字符串传递给javascript函数的问题

2 个答案: