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php - 通过 MySQL 动态启用和禁用 DIV

转载 作者:行者123 更新时间:2023-11-29 13:19:16 25 4
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昨晚,我试图弄清楚如何在提交按钮上方、最底部的联系表单上动态启用和禁用 span#txtCaptchaDiv

因此,我向 MySQL 添加了一个名为 captcha 的新字段,我希望其中 1 显示,0 隐藏

因此,如果我将 1 添加到字段 captcha,以下代码将显示在我的 form.php

 <label for="code">Write code below > <span id="txtCaptchaDiv" style="color:#F00"></span><!-- this is where the script will place the generated code --> 
<input type="hidden" id="txtCaptcha" /></label><!-- this is where the script will place a copy of the code for validation: this is a hidden field -->
<input type="text" name="txtInput" id="txtInput" size="30" />

如果我将 0 添加到字段 captcha,则我的 form.php 上的验证码区域将为空白。

你能帮我一下吗?

这是我当前拥有的index.php代码:

<?php

require_once("/config/database.php");
$con = mysql_connect($config["db_server"],$config["db_user"],$config["db_pass"]);

?>


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Email FORM</title>
</head>

<body>


<div style="width: 550px; text-align: center;">
<span style="filter:alpha(opacity=60); opacity:.6; padding-left: 10px;"><br />


<?php
$data = mysql_query("SELECT * FROM formrelated")
or die(mysql_error());
while($info = mysql_fetch_array( $data ))
Print " ".$info['welcomemsg'] . "";
?>

</span></div>


<form id="form1" name="form1" method="post" action="submit.php" onsubmit="return checkform(this);">
<table width="454" border="1" align="center" cellpadding="0" cellspacing="0">
<tr>
<td width="123">Name</td>
<td width="325">
<input name="name" type="text" />
</td>
</tr>
<tr>
<td height="21">Address</td>
<td><input name="adress" type="text" /></td>
</tr>
<tr>
<td height="21">&nbsp;</td>
<td><input name="address2" type="text" /></td>
</tr>
<tr>
<td height="21">Email</td>
<td><input name="email" type="text" /></td>
</tr>
<tr>
<td height="21">Tel</td>
<td><input name="email" type="text" /></td>
</tr>
</table>

<!--- captcha code here--->

<center>
<table width="454" height="122" border="0" cellspacing="0" cellpadding="0" background="reCAPbg.png">
<tr>
<td height="73" colspan="2" align="center" valign="middle"><label for="code"><span id="txtCaptchaDiv" style="color:#333; font-size:18px;"></span><!-- this is where the script will place the generated code -->
<input type="hidden" id="txtCaptcha" /></label></td>
<td width="136" rowspan="2">&nbsp;</td>
</tr>
<tr>
<td width="145"> type the code here:</td>
<td width="173" height="47" align="center"><input type="text" name="txtInput" id="txtInput" size="20" /></td>
</tr>
</table>
</center>
<!--- captcha code ends here--->

<input name="Submit" type="button" value="submit" />
</form>


<script type="text/javascript">
//Generates the captcha function
var a = Math.ceil(Math.random() * 9)+ '';
var b = Math.ceil(Math.random() * 9)+ '';
var c = Math.ceil(Math.random() * 9)+ '';
var d = Math.ceil(Math.random() * 9)+ '';
var e = Math.ceil(Math.random() * 9)+ '';

var code = a + b + c + d + e;
document.getElementById("txtCaptcha").value = code;
document.getElementById("txtCaptchaDiv").innerHTML = code;
</script>




<script type="text/javascript">
function checkform(theform){
var why = "";

if(theform.txtInput.value == ""){
why += "- Security code should not be empty.\n";
}
if(theform.txtInput.value != ""){
if(ValidCaptcha(theform.txtInput.value) == false){
why += "- Security code did not match.\n";
}
}
if(why != ""){
alert(why);
return false;
}
}

// Validate the Entered input aganist the generated security code function
function ValidCaptcha(){
var str1 = removeSpaces(document.getElementById('txtCaptcha').value);
var str2 = removeSpaces(document.getElementById('txtInput').value);
if (str1 == str2){
return true;
}else{
return false;
}
}

// Remove the spaces from the entered and generated code
function removeSpaces(string){
return string.split(' ').join('');
}

</script>

</body>
</html>

最佳答案

这对你有用......享受吧!

    <?PHP

$query = mysql_query("SELECT captcha FROM formrelated WHERE id = '1'");

while ($row = mysql_fetch_assoc($query)) {
$captchathis = $row['captcha'];

if ($captchathis == "1") {
echo "YOUR HTML CODE HERE";
}
else {
echo "BLANK";
}
}
?>

关于php - 通过 MySQL 动态启用和禁用 DIV,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21066095/

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