好吧，伙计们，基本上我想问的是:我的网站上有一个个人资料页面。现在，用户输入名字和姓氏。例如，稍后他们想更新他们的名字。那么，单个查询中的 UPDATE 语法会更新该 id 的整个字段，对吗？
我想做的是保留姓氏的原始值，即使表更新了名字。我想这样的东西会起作用。
$database = new Database;
if( isset($_POST['submit']) ) {
$id = isset($_POST['user_id'])?$_POST['user_id']:null;
$fname = isset($_POST['fname'])?$_POST['fname']:null;
$lname = isset($_POST['lname'])?$_POST['lname']:null;
$profession = isset($_POST['profession'])?$_POST['profession']:null;
$phone = isset($_POST['phone'])?$_POST['phone']:null;
$fax = isset($_POST['fax'])?$_POST['fax']:null;
$filtered_email = isset($_POST['email'])?$_POST['email']:null;
$workbio = isset($_POST['workbio'])?$_POST['workbio']:null;
$employers = isset($_POST['employers'])?$_POST['employers']:null;
$years = isset($_POST['years_in_industry'])? $_POST['years_in_industry']:null;
$database->query('UPDATE users
SET firstname = COALESCE(:fname,firstname),
lastname = COALESCE(:lname,lastname),
profession = COALESCE(:profession,profession),
phone = COALESCE(:phone,phone),
fax = COALESCE(:fax,fax),
email = COALESCE(:email,email),
projects = COALESCE(:workbio,projects),
companies = COALESCE(:employers,companies),
exp_years = COALESCE(:years_in_industry,exp_years)
WHERE user_id = :id');
$database->bind(':id', $id);
$database->bind(':fname', $fname);
$database->bind(':lname', $lname);
$database->bind(':profession', $profession);
$database->bind(':phone', $phone);
$database->bind(':fax', $fax);
$database->bind(':email', $filtered_email);
$database->bind(':workbio', $workbio);
$database->bind(':employers', $employers);
$database->bind(':years_in_industry', $years);
$database->execute();
}
这些是我的数据库字段名称:user_id、名字、姓氏、职业、电话、传真、电子邮件、项目、公司、exp_years、acct_date
这是我的 html 表单:
<div class="content">
<!-- You only need this form and the form-basic.css -->
<form class="form-basic" method="post" action="<?php $_SERVER['PHP_SELF']; ?>">
<div class="form-title-row">
<h1>Profile:</h1>
</div>
<div class="form-row">
<label>
<span>id:</span>
<input type="hidden" name="id">
</label>
</div>
<div class="form-row">
<label>
<span>Profession:</span>
<input type="text" name="profession">
</label>
</div>
<div class="form-row">
<label>
<span>First name</span>
<input type="text" name="fname">
</label>
</div>
<div class="form-row">
<label>
<span>Last name</span>
<input type="text" name="lname">
</label>
</div>
<div class="form-row">
<label>
<span>Email:</span>
<input type="email" name="email">
</label>
</div>
<div class="form-row">
<label>
<span>Phone:</span>
<input type="tel" name="tel">
</label>
</div>
<div class="form-row">
<label>
<span>Fax:</span>
<input type="tel" name="fax">
</label>
</div>
<div class="form-row">
<label>
<span>Companies I have worked with:</span>
<textarea name="employers"></textarea>
</label>
</div>
<div class="form-row">
<label>
<span>Projects I've worked on:</span>
<textarea name="workbio"></textarea>
</label>
</div>
<div class="form-row">
<label>
<span>Years I've been in my industry:</span>
<select name="years_in_industry">
<option>1 year</option>
<option>2+ years</option>
<option>5+ years</option>
<option>10+ years</option>
</select>
</label>
</div>
<div class="form-row">
<label><span>Radio</span></label>
<div class="form-radio-buttons">
<div>
<label>
<input type="radio" name="radio">
<span>Radio option 1</span>
</label>
</div>
<div>
<label>
<input type="radio" name="radio">
<span>Radio option 2</span>
</label>
</div>
<div>
<label>
<input type="radio" name="radio">
<span>Radio option 3</span>
</label>
</div>
</div>
</div>
<div class="form-row">
<button type="submit">Submit Form</button>
</div>
</form>
谢谢。
最佳答案
如果我理解正确，您可以仅使用 UPDATE 命令来执行此操作。
if( isset($_POST['submit']) ) {
$firstname = $_POST['fname'];
$lastname = $_POST['lname'];
$db->query('UPDATE users
SET fname = COALESCE(:firstname,fname),
lname = COALESCE(:lastname,lname)
WHERE id = :id');
$db->bind(':fname', $firstname);
$db->bind(':lname', $lastname);
$db->bind(':id', $_POST['id']); //it seems you forget the ID
$db->execute();
}
使用COALESCE函数，您只能在字段不为空时更改字段的值，如果为空，则将使用数据库中的相同值。
UPDATE 只会更新您在 SET 子句中指定的字段，所有其他字段将保持相同的值。
您的代码实际上只是缺少 ID 值的绑定(bind)。它实际上会按照你想要的方式工作。我建议的代码是一种改进，只运行一个更新命令而不是两个。
编辑
其实我犯了一个小错误。如果没有这样的索引，则 _POST 变量将计算为错误，假设您尝试了 $_POST['someVar'] 但 someVar 在您的表单中不存在，它将返回错误未定义索引
因此，为了解决这个问题，您可以将当前代码编辑为:
if( isset($_POST['submit']) ) {
$id = isset($_POST['user_id'])?$_POST['user_id']:null;
$fname = isset($_POST['fname'])?$_POST['fname']:null;
$lname = isset($_POST['lname'])?$_POST['lname']:null;
$profession = isset($_POST['profession'])?$_POST['profession']:null;
$phone = isset($_POST['phone'])?$_POST['phone']:null;
$fax = isset($_POST['fax'])?$_POST['fax']:null;
$filtered_email = isset($_POST['email'])?$_POST['email']:null;
$workbio = isset($_POST['workbio'])?$_POST['workbio']:null;
$employers = isset($_POST['employers'])?$_POST['employers']:null;
$years = isset($_POST['years_in_industry'])?$_POST['years_in_industry']:null;
这是一个三元运算符，我在其中检查该变量是否来自您的表单，如果不是，我将 null 分配给它。三元条件是 condition?value if true:value if false;
使用此代码，您的更新命令应该与 coalesce 命令一起使用。
关于php - 部分更新 sql 表并保留未编辑的值的好做法是什么？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/38294512/