Question

我已经编写了这段代码，但它不起作用。我很确定它大部分是正确的，但我不知道它有什么问题。脚本返回输出“undefined”

------ PHP ON SERVER -----------------（http://www.autofficinacicco.it/json.php）

<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");

$conn = new mysqli($nomehost, $nomeuser, $password, "");

$result = $conn->query("SELECT codice, immagine, testo FROM Promozioni");

$outp = array();
 while($rs = $result->fetch_array(MYSQLI_ASSOC)) 
 {
     $outp[]=$rs;    
 }


$conn->close();

echo json_encode($outp);
?>

----- JAVA SCRIPT ----------------------------------------

<script>
var xmlhttp = new XMLHttpRequest();
 var url = "http://www.autofficinacicco.it/json.php";

 xmlhttp.onreadystatechange=function() {
    if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
        myFunction(xmlhttp.responseText);
    }
}
 xmlhttp.open("GET", url, true);
xmlhttp.send();

function myFunction(response) {
    var arr = JSON.parse(response);
    var i;
    var out = "<table>";

    for(i = 0; i < arr.length; i++) {
         out+=arr["immagine"];
         out+="</br>";
         out+=arr["codice"];
         out+="</br>";
         out+=arr["testo"];
         out+="</br>";
    }
    out += "</table>"
    document.getElementById("id").innerHTML =out;
}
</script>

Answer 1

您的JSON不是正确的文档。更改json格式并使用json_encode（）;功能

$outp = "[";
while($rs = $result->fetch_array(MYSQLI_ASSOC)) 
{
 if ($outp != "[") {$outp .= ",";}
 $outp .= '{"Codice":"'  . $rs["codice"] . '",';
 $outp .= '"Testo":"'   . $rs["testo"]        . '",';
 $outp .= '"Immagine":"'. $rs["immagine"]     . '"}'; 
}
$outp .="]";

$conn->close();

到

$outp = array();
while($rs = $result->fetch_array(MYSQLI_ASSOC)) 
{
  $outp[] = $rs;
}
$conn->close();
echo json_encode($outp);

使用Javascript读取JSON对象（错误）

1 个答案: