我有一个名为property_amenities的表，其中包含列; amenity_id和amenity_name，其值在表单上填充为复选框。表单提交后，选中的值将作为数组插入列property_listings的{{1}}表中。
property_amenities上面的代码工作正常。唯一的问题是，当我提交表单并且页面重新加载 <?php
$db_host="localhost";
$db_name="cl43-realv3";
$db_user="root";
$db_pass="";
try
{
$DB_con = new PDO("mysql:host={$db_host};dbname={$db_name}",$db_user,$db_pass);
$DB_con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(PDOException $e)
{
$e->getMessage();
}
if (isset($_POST['list_property'])) {
if (isset($_POST['property_amenities'])) {
$property_amenities = implode(",",$_POST['property_amenities']);
}else{ $property_amenities = "";}
}
?>
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="post">
<div class="row clearfix">
<div class="col-lg-12 col-md-6 col-sm-12">
<?php
$stm = $DB_con->prepare("SELECT * FROM property_amenities");
$stm->execute(array(
));
while ($row = $stm->fetch()){
$amenity_id = $row['amenity_id'];
$amenity_name = $row['amenity_name'];
$List[] ="<label class=checkbox> <input type=checkbox name='property_amenities[]' value='$amenity_name'> $amenity_name</label>";
}
$allamenities = $stm->rowCount();
$how_many_chunks = $allamenities/3;
$roster_chunks = array_chunk($List, round($how_many_chunks), true);
?>
<label>Property Amenities</label></small>
<?php
foreach ($roster_chunks as $roster_chunk_key => $roster_chunk_value) {
echo "<div class='col-lg-3 col-md-6 col-sm-12'>";
echo join($roster_chunk_value);
echo '</div>';
}
?>
</div>
</div><!-- end row -->
<button type="submit" class="btn btn-primary" name="list_property" >SUBMIT PROPERTY</button>
</form>
</body>
</html>
的所有复选框都未选中时，我想要的是在页面提交之前检查的复选框。
我怎样才能做到这一点？
这是property_amenities表
property_amenities答案 0 :(得分：2)
只需在您的数据库中检查选中了哪些复选框，并将<div class="outer">
<div class="inner">Click me</div>
</div>属性包含在您回复复选框时的那些属性中：
checked修改
（回应comment）
由于每个$checked = (condition) ? "checked" : "";
$List[] = "<label class=checkbox>
<input type=checkbox name='...' value='$amenity_name' $checked/>
$amenity_name
</label>";
的{{1}}是便利设施的名称，因此您可以使用提交表单时获得的value数组轻松找到复选框：
checkbox）创建一个空数组。$_POST['property_amenities']检查是否有任何经过检查的便利设施。如果是，请更新上面创建的数组的值。$amenities_checked循环中，检查该虚拟设施的名称是否在该数组中。步骤1＆amp;的代码2：
isset($_POST["property_amenities"])第3步的代码：
while完整代码：
（该代码段用于折叠代码）
# Initialise an array to store the amenities checked.
$amenities_checked = [];
# Check whether the form has been submitted.
if (isset($_POST["list_property"])) {
# Initialise the amenities string.
$property_amenities = "";
# Check whether any checkbox has been checked.
if (isset($_POST["property_amenities"])) {
# Update the checked amenities array.
$amenities_checked = $_POST["property_amenities"];
# Implode the array into a comma-separated string.
$property_amenities = implode(",", $amenities_checked);
}
}
答案 1 :(得分：0)
首先，您从property_listings表中获取数据，并使用explode函数将property_amenities字符串转换为数组。然后你可以使用in_array（）函数来检查你的字符串。如果数组中存在amenity_name，则可以使用checked属性。
例如
$stm = $DB_con->prepare("SELECT * FROM property_amenities");
$stm->execute(array());
// fetch data from property_amenities then use explode function
$property_amenities = explode(",",$data);
while ($row = $stm->fetch()){
$amenity_id = $row['amenity_id'];
$amenity_name = $row['amenity_name'];
$List[] ="<label class=checkbox> <input type=checkbox name='property_amenities[]' value='$amenity_name' <?php if(in_array($amenity_name,$property_amenities)){ echo "checked=true"; }?>> $amenity_name</label>";
}