Question

我正在尝试使用php回显javascript代码并尝试在这个非常回显的代码中从我的数据库中检索数据...有人知道我做错了什么吗？是因为javascript只是客户端吗？（从技术上讲，我正在尝试使用php访问数据库......）我将非常感谢您的帮助！

<?php
if($condition == true){

//connect to the database
//-select the database to use
//-query the database table
//-run the query against the mysql query function
//-create while loop and loop through result set 



echo    "<script>  
    items_set = [
        {
         src : '<?php echo ".$row['imageURL']." ?>', 
         url : '<?php echo ".$row['URL']." ?>', 
         category: '<?php echo ".$row['DetailCategory']." ?>', 
         title : '<?php echo ".$row['Name']." ?>', 
         description : '<?php echo ".$row['Description']." ?>', 
         price : '<?php echo ".$row['Price']." ?>', 
         location : '<?php echo ".$row['Postcode']." ?>', 
         thirdparty : '<?php echo ".$row['ThirdParty']." ?>', 
         thirdparty_mobile : '<?php echo ".$row['Thirdparty']." ?>'
        }           
    ];
    jQuery('#list').portfolio_addon({
        load_count : 1,
        items : items_set
    }); 
</script>";


}}

?>

提前致谢！（在Dreamweaver中，代码的颜色看起来正确，不像这里）

Answer 1

确保将服务器中执行的内容（PHP）与传递给客户端的内容分开并在浏览器中执行：

<?php if(condition == true){ ?>
<script>
    items_set = [
        {src : "<?php echo $row['imageURL']; ?>", 
            url : "<?php echo $row['URL']; ?>", 
            category: "<?php echo $row['DetailCategory']; ?>",
            title : "<?php echo $row['Name']; ?>",
            description : "<?php echo $row['Description']; ?>", 
            price : "<?php echo $row['Price']; ?>",
            location : "<?php echo $row['Postcode']; ?>", 
            thirdparty : "<?php echo $row['ThirdParty']; ?>",
            thirdparty_mobile : "<?php echo $row['Thirdparty']; ?>"}         
    ];
    jQuery('#list').portfolio_addon({
        load_count : 1,
        items : items_set
    }); 
</script>
<?php } ?>

请注意，我修改了代码以生成字符串变量，因为$row变量是PHP代码的一部分，而不是JavaScript。

Answer 2

在你的php中：

//specify header..

header("content-type:application/json");

//create variables for your values

 $row['values'];

//create oo array
arrayResult = 
  array(
 array("variable-name" => $variable,"imageURL" => $imageURL)
      );

在你的HTML ..

    //get data using JSON
    <script type="text/javascript">
    $j.getJSON( "cellJson.php", function( json ) {
        $j.each(json, function(i, item) {
    //example
    var imageURL = item.imageURL;

    //then in your script

      items_set = [
            {
src : ''+imageURL+'', 
url : ''+URL+'', 
category: ''+DetailCategory+'', 
title : ''+Name+''
            }         
        ];
        jQuery('#list').portfolio_addon({
            load_count : 1,
            items : items_set
        });

    });

    </script>

用php回应javascript，包括mysql行变量

2 个答案: