我正在编写一个将发布到 mySQL 数据库的 php 应用程序。我有那部分工作，但我有一个名为 repID 的列，它将包含客户端修复 ID。
将内容添加到数据库的方式是通过我网站上的管理员。我想这样做，如果技术人员输入现有的 repID，它只会更新 key 。而不是制作具有不同修复状态的副本
PHP
$repID = mysql_real_escape_string($_POST['repID']);
$clientName = mysql_real_escape_string($_POST['clientName']);
$devModel = mysql_real_escape_string($_POST['devModel']);
$repStatus = mysql_real_escape_string($_POST['repStatus']);
$tracking = mysql_real_escape_string($_POST['tracking']);
$sql = mysql_query("INSERT INTO status (`repID`, `clientName`, `devModel`, `repStatus`, `tracking`) VALUES ('$repID','$clientName','$devModel','$repStatus', '$tracking');");
输入页面
<?php
include_once('../resources/init.php');
$query = "SELECT * FROM status WHERE repStatus != 'Finished';";
$result = mysql_query($query);
$num = mysql_numrows($result);
mysql_close();
$random = rand(1000000000, 9999999999);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd">
<html>
<head>
<title>Add A Repair</title>
</head>
<body>
<div id="wrapper">
<div id="application">
<div id="randomNum">
<?php echo $random; ?>
</div>
<form method="post" action="insert.php">
<div id="repID">
<label for="repID">Repair ID</label>
<input type="text" name="repID" />
</div>
<div id="clientName">
<label for="clientName">Client Name</label>
<input type="text" name="clientName" />
</div>
<div id="devModel">
<label for="devModel">Device Model</label>
<input type="text" name="devModel" />
</div>
<div id="repStatus">
<label for="repStatus">Repair Status</label>
<select name="repStatus">
<option value="Diagnosis Stage">Diagnosis Stage</option>
<option value="Problem Found">Problem Found</option>
<option value="Possible Solution">Possible Solution</option>
<option value="Parts Ordered">Parts Ordered</option>
<option value="Parts Recieved">Parts Recieved</option>
<option value="Parts/Software Installation Stage">Parts/Software m Installation Stage</option>
<option value="Testing Stage">Testing Stage</option>
<option value="Finished">Finished</option>
</select>
</div>
<div id="tracking">
<label for="tracking">Tracking Number</label>
<input type="text" name="tracking" />
</div>
<div id="submit">
<input type="submit" value="Submit" />
</div>
</form>
<div id="currentClients">
Current Clients
<br /><br />
<table border="0" cellspacing="2" cellpadding="2">
<tr>
<th>Repair ID</th>
<th>Client Name</th>
<th>Device Model</th>
<th>Repair Status</th>
<th>Tracking</th>
</tr>
<?php
$i = 0;
while ($i < $num) {
$v1 = mysql_result($result, $i, "repID");
$v2 = mysql_result($result, $i, "clientName");
$v3 = mysql_result($result, $i, "devModel");
$v4 = mysql_result($result, $i, "repStatus");
$v5 = mysql_result($result, $i, "tracking");
?>
<tr>
<td><?php echo $v1; ?></td>
<td><?php echo $v2; ?></td>
<td><?php echo $v3; ?></td>
<td><?php echo $v4; ?></td>
<td><?php echo $v5; ?></td>
</tr>
<?php
$i++;
}
?>
</table>
</div>
</div>
</div>
</body>
</html>
最佳答案
使用 MySQL 的 INSERT ... ON DUPLICATE KEY UPDATE .当然要确保 repID 是主键或唯一键。
关于php - 如何在重复的列值上更新 mySQL 键，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7462928/