如果状态为非活动状态,如何避免行内容中的删除线? [英] How to avodi the strike-through in the row content if the status is inactive?
问题描述
如果状态为非活动状态,我有一个问题要避免行内容$row['filing_code_refer']
成为删除线,我的代码向我展示了下面的输出不能避免$row['filing_code_refer']
成为删除线.
I have a problem to avoid the row content $row['filing_code_refer']
become the strike-through if the status is inactive, my coding show me the output below cannot avoid the $row['filing_code_refer']
become the strike-through.
下面是我的编码:
<?php
$folderData = mysqli_query($mysql_con,"SELECT * FROM filing_code_management");
// $arr_sql5 = db_conn_select($folderData);
// foreach ($arr_sql5 as $rs_sql5) {
// $active = $rs_sql5['status'];
// }
$folders_arr = array();
while($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'>". $row['category']."</span>" . ' ' . "Rujuk Kod" .$row['filing_code_refer'],
"category" => $row['category'],
"filing_code_refer" => $row['filing_code_refer'],
// "status" => $row['status'], // status 0 is inactive, status 1 is active
"data" => array("status" => $row['status']) ,
"state" => array("selected" => $selected,"opened"=>$opened)
);
}
?>
<!-- Initialize jsTree -->
<div id="folder_jstree" title=""></div>
<!-- Store folder list in JSON format -->
<textarea style="" id='txt_folderjsondata'><?= json_encode($folders_arr) ?></textarea>
<script style="text/javascript">
$(document).ready(function() {
var folder_jsondata = JSON.parse($('#txt_folderjsondata').val());
$('#folder_jstree').jstree({
'core': {
'data': folder_jsondata,
'multiple': false
},
'plugins': ['sort'],
'sort': function(a, b) {
return this.get_text(a).localeCompare(this.get_text(b), 'en', {
numeric: true
});
}
});
var getColor = function(i) {
if (i >= 100 && i <= 199) {
return "blue";
} else if (i >= 200 && i <= 299) {
return "red";
} else if (i >= 300 && i <= 399) {
return "yellow";
} else if (i >= 400 && i <= 499) {
return "purple";
} else if (i >= 500 && i <= 599) {
return "green";
} else {
return "#000";
}
};
var colorNodes = function(nodelist) {
var getStrike = function(status) {
if (status === "0") {
return "line-through;";
} else {
return "";
}
};
var tree = $('#folder_jstree').jstree(true);
nodelist.forEach(function(n) {
tree.get_node(n.id).a_attr.style = "color:" + getColor(parseInt(n.text.substr(0, 3), 10))+ ";"+"text-decoration:" + getStrike(n.data.status);
tree.redraw_node(n.id); //Redraw tree
colorNodes(n.children); //Update leaf nodes
});
};
$('#folder_jstree').bind('load_node.jstree', function(e, data) {
var tree = $('#folder_jstree').jstree(true);
colorNodes(tree.get_json());
});
$('#folder_jstree').bind('hover_node.jstree', function(e, data) {
$("#" + data.node.id).attr("title", data.node.original.category);
});
});
</script>
这是我正在工作的JSfiddle: https://jsfiddle.net/ason5861_cs/3agdhkqf/3/
This is my working JSfiddle : https://jsfiddle.net/ason5861_cs/3agdhkqf/3/
实际上,如果文件夹名称状态为inactive(0),我希望图片下方的输出可以显示filing_code_refer
.如果处于活动状态,则不会显示filing_code_refer
内容.希望有人可以指导我如何解决它.谢谢.
Actually, I want the output like below the picture can show filing_code_refer
if the folder name status is inactive(0). If active will no show the filing_code_refer
content.Hope someone can guide me how to solve it. Thanks.
推荐答案
在HTML 4和XHTML 1中已弃用此元素,而在HTML5中已弃用该元素.如果在语义上适当,即表示已删除的内容,请改用.在所有其他情况下,请使用.
This element is deprecated in HTML 4 and XHTML 1, and obsoleted in HTML5. If semantically appropriate, i.e., if it represents deleted content, use instead. In all other cases use .
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