[英]How to return array in php to $.ajax success function
我想將json數組返回給調用$.ajax function ，但是我只得到了預期數組的最后一項。 也許我不產生數組？
如果我單擊ID為“ btn_getAnswers”的按鈕，則會觸發“ "$("#btn_getAnswers").click" ，並執行代碼"DBCOMANSWERS" 。 我希望"$result" DBCOMANSWERS”中的"$result"是一個填充有我的MYSQL數據庫值的數組。 我返回格式為JSON的"$result" 。 返回的結果應附加到id為“ output”的段落中。 到目前為止，一切正常，但是我除了要返回的三個字符串並附加到該段落之外，現在僅附加了一個，即數據庫中最后一個捕獲的條目。
我真的看不到我必須在哪里添加附加循環或其他內容。 返回的$ result可能不只是數組的最后一個條目，因為它會被覆蓋嗎？
Index.html：
<!DOCTYPE html>
<html>
<head>
<script src="jquery-1.12.3.js"></script> <!-- Import the jquery extension -->
<script>
$(document).ready(function () {
$("#btn_getQuestion").click(function () {
$.ajax({
type: "POST",
url: "DBCOMQUESTIONS.php?q=" + $("#input").val(),
success: function (result) { //Performs an async AJAX request
if (result) {
$("#output").html(result); //assign the value of the result to the paragraph with the id "output"
}
}
});
});
$("#btn_getAnswers").click(function () {
$.ajax({
type: "POST",
url: "DBCOMANSWERS.php?q=" + $("#input").val(),
success: function (result) { //Performs an async AJAX request
if (result) {
$("#output").append(result);
}
}
});
});
});
</script>
</head>
<body>
<p id="output">This is a paragraph.</p>
<input id="input"/>
<button id="btn_getQuestion">Question</button>
<button id="btn_getAnswers">Answers</button>
</body>
</html>
DBCOMANSWERS.php：
<!DOCTYPE HTML>
<head>
</head>
<body>
<?php
include("connection.php"); //includes mysqli_connent with database
include("ErrorHandler.php"); //includes error handling function
set_error_handler("ErrorHandler"); //set the new error handler
$q = intval($_GET['q']);
$sql="SELECT * FROM tbl_answers WHERE QID ='".$q."'"; //define sql statement
$query = mysqli_query($con,$sql); // get the data from the db
while ($row = $query->fetch_array(MYSQLI_ASSOC)) { // fetches a result row as an associative array
$result = $row['answer'];
}
echo json_encode($result); // return value of $result
mysqli_close($con); // close connection with database
?>
</body>
<html>
嘗試：刪除所有html標簽，然后
include("ErrorHandler.php"); //includes error handling function
set_error_handler("ErrorHandler"); //set the new error handler
從ajaxed php文件中，創建結果數組並將每個結果附加到該數組
$result = []
while ($row = $query->fetch_array(MYSQLI_ASSOC)) { // fetches a result row as an associative array
$result[] = $row['answer'];
}
header('Content-Type: application/json');//change header to json format
在您的ajax函數中，您需要執行一個循環：
success: function(result){ //Performs an async AJAX request
result.forEach(function(i,v){
$("#output").append(v.answer);
})
}}
你需要做兩件事
刪除html並添加數組集合。 這就是您的DBCOMANSWERS.php必須看起來像的樣子
<?php
include("connection.php"); //includes mysqli_connent with database
include("ErrorHandler.php"); //includes error handling function
set_error_handler("ErrorHandler"); //set the new error handler
$q = intval($_GET['q']);
$sql="SELECT * FROM tbl_answers WHERE QID ='".$q."'"; //define sql statement
$query = mysqli_query($con,$sql); // get the data from the db
$result = [];
while ($row = $query->fetch_array(MYSQLI_ASSOC)) { // fetches a result row as an associative array
$result [] = $row['answer'];
}
mysqli_close($con); // close connection with database
header('Content-Type: application/json');
echo json_encode($result); // return value of $result
?>
然后在您的html中，如@madalinivascu建議
success: function(result){ //Performs an async AJAX request
result.forEach(function(i,v){
$("#output").append(v.answer);
})
}}
嘗試：
$result = []
while ($row = $query->fetch_assoc()) { // fetches a result row as an associative array
$result[] = $row['answer'];
}
參考：
http://php.net/manual/en/mysqli-result.fetch-array.php
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