Question

以下是我的AJAX代码。我在评论中写了这些问题。

$.ajax({
type: "POST",
url: "dbtryout2_2.php",
data: datastr,
success: function(arrayphp) {
     //here one by one text is being received from php  from within a loop
     //each text item is getting displayed as link here
     //it is OK that text is getting displayed as link
     //BUT the problem is that all the text returned from php are getting displayed as 
     //one link i.e between one "<a></a>"
     //I WANT THAT EACH TEXT SHOULD BE DISPLAYED AS SEPERATE LINKS
     //AS EACH TEXT IS GETTING RETURNED FROM WITHIN SEPARATE ITERATIONS OF PHP LOOP
     var link = $('<a href="#" class="album"><font color="red">' + arrayphp + '</font></a>');
    linkclass = link.attr("class"); 
    $(".searchby .searchlist").append(link);   
}
});

怎么办？

Answer 1

Php将所有内容立即抛给ajax成功函数...你需要在php数组上使用json_encode并回显它以将其作为json数组传递。

<?php 
$return = array();
foreach($array as $key=>$value){
 //Do your processing of $value here
 $processed_value = $value;
 array_push($return, $processed_value);
}
echo json_encode($return);
?>

在ajax调用中将dataType设置为“json”，以便将从php传递的数据解析为json数组。使用每个jquery遍历接收到的数组。

$.ajax({
type: "POST",
url: "dbtryout2_2.php",
data: datastr,
dataType: "json",
success: function(arrayjson) {
     $.each(function(arrayjson, function(i, processed_value) {
       var link = $('<a href="#" class="album"><font color="red">' + processed_value + '</font></a>');
       linkclass = link.attr("class"); 
       $(".searchby .searchlist").append(link); 
    });  
}
});

显示从循环中的php返回的文本作为html页面上的单独链接

1 个答案: