[英]PHP script not returning query, but the MySql query works
我的最终目标是让js脚本将php查询的结果作为json返回,以便我可以使用它进行恶意操作。
我有要使用的MySql查询,当我在Workbench中对其进行测试时,它肯定可以正常工作,但是当我在PHP脚本中对其进行尝试时,什么也没有返回。 PHP中的当前查询只是一个占位符,用于测试数据交换。
我手动将一些东西放到数组(searchResults)中,并且得到返回,但是NOTHING来自执行该脚本的PHP脚本。
我也知道这可能不是最安全或最有效的代码,我只是想让它在此时工作。
<?php
// Include your database creds and login to the db
require_once 'login_karavites.php';
$db = mysqli_connect($db_hostname, $db_username, $db_password);
// Handle the input/request.
$searchString_UNSAFE = $_POST['eName']; // change that, obviously
// Bare minimum sanitation to prevent injection.
$searchString = $db->escape_string($searchString_UNSAFE);
// Construct the SQL query
$sql = "SELECT * FROM `Halls` WHERE hall_name = 'Rose Ballroom'";
// Do the database lookup.
$result = $db->query($sql);
// Create empty array to hold our results (to be sent back to the browser).
$searchResults = array();
$searchResults[]="wow";
// If we had results, put them into that array
if ($result->num_rows > 0) {
// This loop will retrieve every row from that result set
while ($row = $result->fetch_assoc()) {
// From each row, just take the 'event_name' field.
$searchResults[] = $row['hall_name'];
}
}
// Done with the db, now we just have to send the results back to the browser.
$db->close();
// Send the correct content-type header.
// This ensures that jQuery automatically converts the response into an
// array or object, rather than just treating it like a block of text.
// Must be the FIRST thing the PHP script outputs, or it will choke.
header('Content-type: application/json');
// Output the data.
echo json_encode($searchResults);
?>
js脚本。
$(document).ready(function() {
// All this stuff runs as soon as the page is fully loaded
// Attach a function to the Submit action on #eventForm
$('#eventForm').submit(function() {
// Submit the form via AJAX
$(this).ajaxSubmit({
// Attach a function to the "the PHP script returned some results" event
success: function(response, status, xhr, $form){
// I am assuming that this is your data format, for example:
// { "searchResults": [ "result1", "result2", "result3" ] }
// I am also assuming that you want your results in div#results
$('div#results').html(""); // Clear it out of anything that's already there.
console.log(response);
for (i in response['searchResults']) {
$('div#results').append( response['searchResults'][i] );
}
},
// Give up if PHP doesn't answer in 3 seconds
timeout: 3000,
// Path to the PHP file we want to send this to
url: 'phpdata/eventsData.php'
});
// Make sure the browser does NOT proceed to submit the form again,
// the old fashioned way (full page reload).
return false;
});
});
我的问题的哪一部分是:
我愚蠢的是从未将表单方法设置为原始表单,因此POST实际上并没有通过。
我在PHP文件中设置了错误的MySql连接。
我自己的笔记,请确保您检查所有设置。
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