我在 AJAX 中的功能有問題。我創(chuàng)建了一個對返回 JSON 的 PHP 文件的 AJAX 調(diào)用。對于循環(huán)此 JSON，我創(chuàng)建了一個函數(shù)，如果 AJAX 成功，我將運行該函數(shù)。但實際上數(shù)據(jù)是空的。 <script> document.getElementById("getproducts").addEventListener("submit", sendAjax); function sendAjax(event) { var q = document.getElementById('search').value; var xhttp = new XMLHttpRequest(); xhttp.onreadystatechange = function() { if (this.readyState == 4 && this.status == 200) { display(this.responseText); } } xhttp.open("POST", "results.php", true); xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); xhttp.send('search='+q); event.preventDefault(); } function display( jsdata ){ for ( var key in jsdata ){ var htmltabel = ''; var datanode = document.createElement("div"); htmltabel += '<div class="id">' + jsdata[key]['id'] + '</div>'; content = htmltabel; datanode.innerHTML = content; document.getElementById("resultt").appendChild(datanode); } } </script>如果我像這樣在函數(shù)中編寫 JSON 硬代碼，那么一切都很好。var hardcoded = {"1736":{"id":"1736","post_title":"Test explode","_sku":"12345","_stock":null,"_price":"9.50"}}//PART OF THE CODEif (this.readyState == 4 && this.status == 200) { display(hardcoded);}我該如何解決該函數(shù)使用響應(yīng)的 JSON 的問題？
1 回答
一只甜甜圈
TA貢獻(xiàn)1836條經(jīng)驗 獲得超5個贊
這是一個更正的腳本，您應(yīng)該只將 responseData 從轉(zhuǎn)換string為Json Object！
document.getElementById("getproducts").addEventListener("submit", sendAjax);
function sendAjax(event) {
var q = document.getElementById('search').value;
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
display( JSON.parse(this.responseText) ); // You should convert the response from string to a valid JSON
}
}
xhttp.open("POST", "results.php", true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send('search='+q);
event.preventDefault();
}
function display( jsdata ){
for ( var key in jsdata ){
var htmltabel = '';
var datanode = document.createElement("div");
htmltabel += '<div class="id">' + jsdata[key]['id'] + '</div>';
content = htmltabel;
datanode.innerHTML = content;
document.getElementById("resultt").appendChild(datanode);
}
}
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