如何根据PHP / Javascript / Mysql中的表列来显示输入类型的条件？ [英] How to do a condition displaying input type based on table column in PHP/Javascript/Mysql?
问题描述
这是我的数据库表。我目前正在将option1-10显示为文本，但是我想将其显示为每个选项的输入类型，具体取决于其answer_type列是单选按钮还是复选框。当option1-10的值为null时，也不应显示。另外，如何将其分组为一个，例如，如果是单选按钮，我应该只能选择一个？
This is my database table. I'm currently displaying option1-10 as text but I want to display it as an input type for each option depending if its answer_type column is radiobutton or checkbox. It also shouldn't display when the value of option1-10 is null. Also how do i group it into one like for example if it's a radiobutton, i should only be able to choose one?
这是我显示文本的代码
ajax
<html>
<head>
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xmlhttp.open("GET","hay.php?q="+str,true);
xmlhttp.send();
}
}
</script>
</head>
</html>
<form>
<select name="users" onchange="showUser(this.value)">
<option>-None Selected-</option>
<?php
$con = mysqli_connect('localhost','root','','imetrics') or die ("Cannot connect");
$query=mysqli_query($con, "SELECT * FROM question");
while($row=mysqli_fetch_array($query)) {
?>
<option value="<?php echo $row["question_id"]; ?>"><?php echo $row["questiontitle"]; ?></option>
<?php
}
?>
</select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b></div>
php代码
<?php
$con = mysqli_connect('localhost','root','','imetrics') or die ("Cannot connect to database");
if(!$con){
echo ('Could not connect: ' . mysqli_error($con));
}
$id= isset($_GET["q"])?$_GET["q"]:"";
$query = mysqli_query($con, "SELECT * FROM question WHERE question_id = '".$id."'");
while($row = mysqli_fetch_array($query)) {
echo $row['Option_1'] . "<br>";
echo $row['Option_2'] . "<br>";
echo $row['Option_3'] . "<br>";
echo $row['Option_4'] . "<br>";
echo $row['Option_5'] . "<br>";
echo $row['Option_6'] . "<br>";
echo $row['Option_7'] . "<br>";
echo $row['Option_8'] . "<br>";
echo $row['Option_9'] . "<br>";
echo $row['Option_10'] . "<br>";
}
?>
What's the right condition for my problem?
推荐答案
您只需要在其中添加更多工作。另外，将intval用于防止SQL注入攻击。我正在使用一个功能来防止复制粘贴代码重复。我还要从数据库转义数据，因为它可能包含<
You just have to put a bit more work into it. Also, using intval for SQL injection attack prevention. I am using a function to prevent duplicate copy-pasted code. I'm also escaping the data from the database as it might contain "<" or "&".
<?php
$con = mysqli_connect('localhost','root','','imetrics') or die ("Cannot connect to database");
if(!$con){
echo ('Could not connect: ' . mysqli_error($con));
}
$id= isset($_GET["q"])?intval($_GET["q"]):"";
$query = mysqli_query($con, "SELECT * FROM question WHERE question_id = '".$id."'");
function displayOption($i, $value, $answer_type) {
if($value == null) {
return;
}
if($answer_type == "radiobutton") {
echo '<input type="radio" name="rinput" value="'.htmlspecialchars($value, ENT_QUOTES).'">'.htmlspecialchars($value).'<br>';
} else if($answer_type == "checkbox") {
echo '<input type="checkbox" name="cinput['.$i.']" value="'.htmlspecialchars($value, ENT_QUOTES).'">'.htmlspecialchars($value).'<br>';
}
}
while($row = mysqli_fetch_assoc($query)) {
for($i = 1; $i<=10; ++$i) {
displayOption($i, $row["Option_$i"], $row['answer_type']);
}
}
?>
这篇关于如何根据PHP / Javascript / Mysql中的表列来显示输入类型的条件？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
