php - 语法错误 : JSON. 解析 : unexpected non-whitespace character after JSON data what is wrong here?

Question

我已经搜索了那个错误并查找了很多帖子..但我仍然无法弄清楚这段代码有什么问题:

我的 ajax 调用:

function myCall3() {
  $.ajax({
    type:"POST",
    url:"ajax3.php",
    dataType:"json",
    success:function(response){
      alert(response[0]);
    }
  });
}

我的 mysql/php 代码:

<?php


// QUERY NEW VINE
$array = array();
$myquery = "SELECT * FROM table1 ORDER BY rand() LIMIT 9";
$result = mysql_query($myquery)
OR die("Error: $myquery <br>".mysql_error());

while($row = mysql_fetch_object($result)){
  $currentid = "$row->id";
  $currentname = "$row->name";
  $currenturl = "$row->url";
  $currentimage = "$row->image";
  $array[] =     array('id'=>$currentid,'url'=>$currenturl,'name'=>$currentname,'image'=>$currentimage);

}

echo json_encode($array);

?>

当我提醒错误时，它说:

SyntaxError: SyntaxError: JSON.parse: unexpected character

Answer 1

您在循环中不断覆盖 $array，然后过早地回显它。

$array =    array('id'=>$currentid,'url'=>$currenturl,'name'=>$currentname,'image'=>$currentimage);

echo json_encode($array);

}

您应该将每一行推送到 $array 变量，然后在循环外进行 json_encode。

$array = array();
$myquery = "SELECT * FROM table1 ORDER BY rand() LIMIT 9";
$result = mysql_query($myquery)
OR die("Error: $myquery <br>".mysql_error());
while($row = mysql_fetch_object($result))
{
$currentid = "$row->id";
$currentname = "$row->name";
$currenturl = "$row->url";
$currentimage = "$row->image";
$array[] =  array('id'=>$currentid,'url'=>$currenturl,'name'=>$currentname,'image'=>$currentimage);

}

echo json_encode($array);

那么你的 javascript 回调将需要一个数组，而不是一个对象，即

console.log(response[0]);

注意:您可以直接将 $row 插入您想要的 $array。