如何通过PHP将复选框数组值存储在MYSQL中? [英] How to store the checkbox array values in MYSQL through PHP?
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问题描述
<div id="project_aminities" class="project_aminities">
<label>Aminities:</label>
<input type="checkbox" value="Gym" name="aminities[]">Gym
<input type="checkbox" value="Indoor Games" name="aminities[]">Indoor Games
<input type="checkbox" value="Children Play Area" name="aminities[]">Children Play Area
<input type="checkbox" value="Swimming Pool" name="aminities[]">Swimming Pool
<input type="checkbox" value="Multi Function Hall" name="aminities[]">Multi Function Hall
<input type="checkbox" value="Jogging Track" name="aminities[]">Jogging Track<br>
</div>
<script>
jQuery(document).ready(function()
{
jQuery('#submit').click(function()
{
var aminities = $('input[type=checkbox]:checked').map(function(_, el) {
return $(el).val();
}).get();
var jsonString = JSON.stringify(aminities);
jQuery.ajax(
{
type:"POST",
url:"sample_project_details.php",
data: {data : jsonString},
cache: false,
success:function(data)
{
alert('OK');
}
});
});
});
</script>
<?php
mysql_connect("localhost","root","myfourwalls");
mysql_select_db("test");
$aminitiesData = json_decode(stripslashes($_POST['data']));
?>
我应该怎样做所选的复选框值存储到MYSQL DB(在名为aminities的列中)?每次提交表单时所选值的数量都不同。
?>
What should I do store the selected checkbox values to MYSQL DB(inside the column called aminities)?Number of selected values differ each time I Submit my Form.
推荐答案
(' input [type = checkbox]:checked')。map(function(_,el){
return
('input[type=checkbox]:checked').map(function(_, el) { return
(el).val();
})。 get ();
var jsonString = JSON.stringify(aminities);
jQuery.ajax(
{
类型: POST,
url: sample_project_details.php,
数据:{data:jsonString},
cache: false ,
成功:函数(数据)
{
alert(' OK');
}
});
});
});
< / script >
(el).val(); }).get(); var jsonString = JSON.stringify(aminities); jQuery.ajax( { type:"POST", url:"sample_project_details.php", data: {data : jsonString}, cache: false, success:function(data) { alert('OK'); } }); }); }); </script>
<?php
mysql_connect("localhost","root","myfourwalls");
mysql_select_db("test");
aminitiesData = json_decode(stripslashes(
aminitiesData = json_decode(stripslashes(
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