【发布时间】:2020-08-07 16:20:29
【问题描述】:
我正在创建 js 树以显示文件夹路径名,我的问题是如何检查数据库表 status 是否为 0(非活动)的条件,然后将在js树。 Else table status 为 1(active) 刚刚恢复正常。以下是我的编码:
<?php
$folderData = mysqli_query($mysql_con,"SELECT * FROM filing_code_management");
$folders_arr = array();
while($row = mysqli_fetch_assoc($folderData)){
$parentid = $row['parentid'];
if($parentid == '0') $parentid = "#";
$selected = false;$opened = false;
if($row['id'] == 2){
$selected = true;$opened = true;
}
$folders_arr[] = array(
"id" => $row['id'],
"parent" => $parentid,
"text" => $row['name'] . ' ' . "<span id='category'>". $row['category']."</span>",
"category" => $row['category'],
"status" => $row['status'], // status 0 is inactive, status 1 is active
"state" => array("selected" => $selected,"opened"=>$opened)
);
}
?>
<script style="text/javascript">
var StrikeNodes = function(nodelist) {
var tree = $('#folder_jstree').jstree(true);
nodelist.forEach(function(n) {
tree.get_node(n.id).a_attr.style = "text-decoration:" + getStrike(parseInt(n.text.substr(0, 3), 10));
tree.redraw_node(n.id); //Redraw tree
StrikeNodes(n.children); //Update leaf nodes
});
};
var getStrike = function(i) {
if (status = '0' ) {
return "line-through;";
} else {
return "";
}
};
$('#folder_jstree').bind('load_node.jstree', function(e, data) {
var tree = $('#folder_jstree').jstree(true);
StrikeNodes(tree.get_json());
});
</script>
现在我的输出显示了 js 树中的所有线路,而不是检测哪个是活动的或不活动的。
我的工作 JSFiddle 代码在这里:https://jsfiddle.net/ason5861_cs/9x0dsotz/3/
希望有人能指导我哪里出错了。
【问题讨论】:
标签: php jquery dom strikethrough