我已经阅读了很多关于堆栈溢出的答案,但我找不到合适的答案。我想从php文件发送多个变量到javascript文件。我想稍后单独使用这些变量。所以请解释一下如何从php文件中获取变量以及如何在以后单独使用它们的简单示例。
这是我的js。
<script>
function here(card_numb) {
alert("pk!");
$.ajax({
url: 'details.php',
type: "GET",
dataType: 'json',
data: ({
card_number: card_numb
}),
success: function(data) {
console.log('card_number:'+data.card_number+'book_issued:'+data.book_isued);
}
});
}
我收到警报'pk!'。但$ .ajax无效。
这是details.php
<?php
if(isset($_GET['card_number'])){
$card_number = $_GET['card_number'];
$query = "Select * from users where card_number = '".$card_number."'";
$query_run = mysqli_query($link,$query);
$row_numb =@mysqli_num_rows($query_run);
if($row_numb == 0){
echo "<div class='bdiv1'>No such number found!</div>";
} else{
$row=mysqli_fetch_assoc($query_run);
$book1 = $row['user_name'];
$arr = array('isued_book' => $book1,'card_number' => $card_number);
echo json_encode($arr);
exit();
}
} ?&GT;
谢谢!
答案 0 :(得分:0)
somthing.js - 你的jspage
<script>
function here(card_numb) {
$.ajax({
url: 'details.php',
type: 'GET',
dataType: 'json',
data: {
card_number: card_numb
},
success: function(data) {
console.log('card_number:'+data.card_number+'book_issued:'+data.isued_book);
}
});
} 成功:功能(结果){ 的console.log( '变量1:' + result.var1 + '变量2:' + result.var2 + 'variable3:' + result.var3); }});
details.php
<?php
if(isset($_GET['card_number'])){
$card_number = $_GET['card_number'];
$query = "Select * from users where card_number = ".$card_number;
$query_run = mysqli_query($link,$query);
$row_numb =@mysqli_num_rows($query_run);
if(!$query_run){
echo "<div class='bdiv1'>No such number found!</div>";
} else {
$row=mysqli_fetch_assoc($query_run);
$book1 = $row['user_name'];
$arr = array('isued_book' => $book1,'card_number' => $card_number);
echo json_encode($arr);
exit();
}
如果当前值进入$ row,您可以在控制台中获得结果