2016-04-29 128 views

回答

2

PHP:

$sql = "SELECT * from reading ORDER BY id DESC LIMIT 1"; 
if ($query = mysqli_query($link, $sql)) { 
    // fetch one result 
    $row = mysqli_fetch_row($query); 
    echo json_encode($row); 
} 

的jQuery:

// sample variables 
var username = ""; 
var password = ""; 
var phpUrl = 'http://sampleurl.com/mypage.php'; 
$.getJSON(phpUrl, function (result) { 
    $.each(result, function (data) { 
     // if data sent back is eg data:[{username:'sample', password:'sample'}] 
     // then you fetch and save as 
     username = data.username; 
     password = data.password; 
    }); 
}); 
0

cookies方式肯定是要走的路。如果你正在做這一切在一個頁面上,只是需要一個快速解決,你可以做

<?php 
$sql = "SELECT * from reading ORDER BY id DESC LIMIT 1"; 
if ($query = mysqli_query($link, $sql)) { 
    // fetch one result 
    $row = mysqli_fetch_row($query); 
    $json_string = json_encode($row); 
} 
?> 

<script> var sql_row = <?php echo $json_string; ?>; </script> 

這是偷懶的辦法,並會導致混亂很快,但它可能對於理解是有用如何PHP和JS一起工作

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