[英]Using Ajax/jQuery to make POST request to insert row to mysql table on node js server
[英]How to disabled table row using JS and insert the chosen option in MYSQL?
我有一个PHP代码段,可动态显示表行。 我每一行都有一个带有“是”和“否”选项的单选按钮。
我创建了一个JS函数,当用户选择一个选项时,将显示一个弹出框。
如果用户在单选按钮中选择“是”选项,然后在弹出框中单击“确定”,则即使单选按钮也将被禁用,表行也将被禁用。 所选的选项将保存在MYSQL中。
如何在MySQL中保存所选选项?
我的禁用行的JS代码段不起作用。 如何解决这个问题?
PHP:
echo '<td id="resumeFile"><a href="' . $dir . $file . '">Download Resume</a></td>';
echo '<td id="radioOption">
<label for="Yes">Yes</label>
<input type="radio" id="processedOptionYes" name="processedOption" value="Yes" onclick="proccessedCheck()"/>
<label for="No">No</label>
<input type="radio" id="processedOptionNo" name="processedOption" value="No" onclick="proccessedCheck()"/></td>';
JS:
function proccessedCheck(){
var checked = null;
var inputs = document.getElementsByName('processedOption');
for (var i = 0; i < inputs.length; i++){
if (inputs[i].checked) {
checked = inputs[i];
break;
}
}
if(checked == null){
return false;
} else if (checked == true){
document.getElementById("resumeFile").disabled = true;
document.getElementById("radioOption").disabled = true;
document.getElementById("resumeFile").title = "This option has been disabled.";
} else {
return confirm('You have chosen '+ checked.value + ', is this correct?');
}
}
好的,如果您要从PHP中回显整个表,只需将参数预置到表中
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<script src="https://code.jquery.com/jquery-2.2.3.min.js" integrity="sha256-a23g1Nt4dtEYOj7bR+vTu7+T8VP13humZFBJNIYoEJo=" crossorigin="anonymous"></script>
<script>
function proccessedCheck(id,answer) {
if (confirm('You have chosen '+ id +': '+ answer + ', is this correct?')) {
$("#processedOptionYes"+id).attr('disabled',true);
$("#processedOptionNo"+id).attr('disabled',true);
var withlink = $("#resumeFile"+id).html();
var withoutlink = $(withlink).html();
$("#resumeFile"+id).html("").append(withoutlink);
$("#input1".val(id);
$("#input2".val(answer);
$("#myform").submit();
}
}
</script>
<!-- EDIT: hidden form to submit -->
<form id="myform" method="POST" action="savedb.php">
<input type="hidden" id="input1" name="id" />
<input type="hidden" id="input2" name="answer" />
</form>
<table>
<tr>
<?php
$dir="";
$file="";
$id = 0;
//foreach($array as $row) {
$id++;
echo '<td id="resumeFile'.$id.'"><a href="' . $dir . $file . '">Download Resume</a></td>';
echo '<td id="radioOption>
<label for="Yes">Yes</label>
<input type="radio" id="processedOptionYes'.$id.'" name="processedOption" value="Yes" onclick="proccessedCheck('.$id.',\'Yes\')"/>
<label for="No">No</label>
<input type="radio" id="processedOptionNo'.$id.'" name="processedOption" value="No" onclick="proccessedCheck('.$id.',\'No\')"/></td>';
//}
?>
</tr>
</table>
</body>
</html>
saveb.php的内容,不必是单独的文件
<?php
// Check if my post array arrived, comment this line when u done
echo "<pre>";print_r($_REQUEST);echo "</pre>"; die();
// Connect to DB
// Build SQL insert string with $_REQUEST['id'] as the primary key
?>
对于初学者,请尝试更换:
document.getElementById("resumeFile").disabled = true;
document.getElementById("radioOption").disabled = true;
与:
document.getElementById("processedOptionYes").disabled = true;
document.getElementById("processedOptionNo").disabled = true;
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