[英]Pass values from JQUERY to PHP then output to INPUT in FORM
我需要从表单获取值,然后使用jquery将其发送到php,然后将结果输出到下拉选择菜单
获取使用jquery的价值
<input id="search" name="search" type="text">
将其发送到php并执行查询
<select id="farmertype" name="farmertype" >
<option value="" > - PLEASE SELECT FARM -</option>
//// output here as options
</select>
我的php文件farm.php
<?php
include_once("../init.php");
$q = ($_POST["search"]);
$db->query("SELECT * FROM farmers ");
while ($line = $db->fetchNextObject()) {
$idno = $line->idno;
echo "<option value='$idno'>$idno</option>";
}
}
?>
jQuery部分是如此混乱,这是我真正需要帮助的地方
$("#search").click(function() {
search = $(this).attr('#search');
$.ajax({
type: 'GET',
url: 'farm.php',
data: "#search=" + search,
});
});
试试这个,它将对您有帮助。
jQuery的:
$("#search").click(function() {
search = $(this).val();
$.ajax({
type: 'POST',
url: 'farm.php',
data: {searchValue:search},
success:function(result) {
console.log(result);
}
});
});
PHP:
<?php
include_once("../init.php");
$q = ($_POST["searchValue"]);
$db->query("SELECT * FROM farmers");
$result = [];
while ($line = $db->fetchNextObject()) {
$idno = $line->idno;
$result = "<option value='$idno'>$idno</option>";
}
print_r($result);
?>
您的变量$ q的用途是什么?
您的jQuery可以像:
$("#search").click(function() {
search = $('#search').val();
$.ajax({
type: 'GET',
url: 'farm.php',
data: {search : search},
success: function(html){
alert(html);
}
});
});
$("#search").click(function() { /* I think you should use keyUp or use click on a button, nobody clicks an input box */
var search = $(this).val();
$.ajax({
method: 'POST', //
url: 'farm.php',
data: {'search' : search},
success: function(data){
alert(data);
}
});
});
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