如何在php中的mysql数据库中插入单个复选框值。
let source = {name:"value",age:"value",linkValue:[[{linkValue:"value1"},{linkValue:"value2"},{linkValue:"value3"},{linkValue:"value4"},{linkValue:"value5"}]]};
// Copy the array in a variable
let yourArray = source.linkValue[0];
// Delete the original array in the source object
delete source.linkValue;
yourArray.forEach((item, index) => {
source["linkValue" + (index + 1)] = item.linkValue
});
console.log(source); // Should have what you want
答案 0 :(得分:1)
男性,女性选择总是无线电,而不是复选框。并且不要将它们的名称作为数组,因为一次只选择其中一个,因此请进行以下更改:
Male <input type="radio" value="male" name="gender" >
Female <input type="radio" value="Female" name="gender">
并获得其值:
$gender = $_REQUEST['gender'];
并使用$gender
答案 1 :(得分:1)
您可以检查其状态并根据它插入
Male <input type="radio" value="male" name="male" >
Female <input type="radio" value="Female" name="female">
在php代码中使用如下的验证:
if(isset($_REQUEST['gender'])) {
$gender = $_REQUEST['gender'];
//Code to insert this $gender in database
}