数据未提交到数据库并显示“错误”消息

Question

有两个php文件包含表单，另一个包含将表单数据插入数据库表中的代码。

这是从第一个文件

<form name="form" action="insert_dataE.php" onSubmit="return 
validation()" method="post" id="formServiceEntry"> 
    <fieldset id="fieldSetServiceEntry"> 
    <legend align="center">Please fill the form</legend> 
    <p class="FieldHeading"><i>Vehicle No(*): </i></p> 
    <input id="VehicleNoFieldArea" type="text" name="VehicleNoField" 
size="6" maxlength="8"/> 
    <p class="FieldHeading"><i>Description(*):</i></p> 
    <textarea id="descriptionFieldArea" name="descriptionField" 
rows="2" cols="20" size="15" maxlength="18"></textarea> 
    <p class="FieldHeading"><i>Total(*):</i></p> 
    <input id="totalFieldArea" name="totalField" type="text" 
size="4" maxlength="4"/> 
    <p id="amountFieldHeading"><i>Bill(*):</i></p> 
    <input id="amountFieldArea" name="amountField" type="text" 
size="3" maxlength="3" onKeyUp="balance();" /> 
    <br/> 
    <div id="divisionRadioButton"> 
     <h3 id="radioButtonHeading">Service(*):</h3> 
     Service &nbsp; 
     <input class="textFields" type="radio" 
name="serviceSelection" value="service" checked /> 
     <br/> 
     Wash &nbsp; 
     <input class="textFields" type="radio" 
name="serviceSelection" value="wash" /> 
    </div> 
    <p id="balanceFieldHeading"><i>Balance(*):</i></p> 
    <input id="balanceFieldArea" name="balanceField" type="text" 
size="4" maxlength="4"/> 
    </fieldset> 
    <input class="btnsSE" type="submit" name="Button" value="Submit" /> 
    <input class="btnsSE" type="reset" name="Button" value="Reset Form"/> 
    <input type="button" class="btnsSE" value="Back to the staff 
interface" onClick="window.location='staffE.php';"/> 
</form> 

这是第二个文件

<?php 
// Connects to your Database 
$conn=mysql_connect("localhost", "webgeek1_service", "6defyu4642070") or 
die(mysql_error()); 
mysql_select_db("webgeek1_software_order", $conn) or die(mysql_error()); 
$result = mysql_query("SELECT * FROM application", $conn); 
$num_rows = mysql_num_rows($result); 
$num_rows = $num_rows + 1; 
$id= $num_rows; 
$dateAndTime = date('y-m-d H:i:s',time()); 
$vehicleNo=mysql_real_escape_string($_POST['VehicleNoField']); 
$description=mysql_real_escape_string($_POST['descriptionField']); 
$amount=mysql_real_escape_string($_POST['amountField']); 
$service=mysql_real_escape_string($_POST['serviceSelection']); 
// Build an sql statment to add the query details 
$sql="INSERT INTO `webgeek1_software_order`.`application`(`serialNo`, 
`dateAndTime` , `vehicleNo` , `description` ,`amount`,`service`) 
VALUES 
('$id',  
'$dateAndTime','$vehicleNo','$description','$amount','$service')"; 
$result = mysql_query($sql, $conn); 
if($result) 
{ 
    echo "<p id='headingInsertData'>Service Station Web Application</p>"; 
    echo "<p id='receiptHeading'>Receipt</p>"; 
    echo "<div id='mainFieldsInsertData'>"; 
    echo "Serial No: " . " " . $id; 
    echo "<br/>"; 
    echo "Date and Time: " . " " . $dateAndTime; 
    echo "<br/>"; 
    echo "Vehicle No: " . " " . $vehicleNo; 
    echo "<br/>"; 
    echo "Description: " . " " . $description; 
    echo "<br/>"; 
    echo "Amount: " . " " . $amount; 
    echo "<br/>"; 
    echo "Service:" . " " . $service; 
    echo "<br/>"; 
    echo "<br/>"; 
    echo"Thanks for using our services"; 
    echo "</div>"; 
    echo "<div id='footerInsertData'>"; 
    echo "<i>Developed by: Web Geeks - Information Technology (IT) 
    Company</i>"; 
    echo "</div>"; 
    echo "<div align='center'>"; 
    echo "<input class='btns' type='button' value='Print' 
    onClick='javascript: window.print();'/>"; 
    echo "<input type='button' class='btns' value='Back to the 
    Application' onClick='newDoc()'/>"; 
    echo "</div>"; 
} 
else 
{ 
    echo "ERROR"; 
} 
// close connection 
mysql_close($conn); 
?> 

Answer 1

您遇到的错误的插入数据的代码提交表单代码（Duplicate entry '51' for key 'PRIMARY'）似乎非常合乎逻辑因为你正在给你指定的ID行而不是自动写入。此外，您正在使用基于当前行数量的ID。这导致了ID重复的MySQL错误。

要解决此问题：

1. 修改serialNo柱和A_I列（AUTO_INCREMENT）勾选复选框。这将确保您始终拥有独特的ID。

2. 完全删除该部分在代码：

$result = mysql_query("SELECT * FROM application", $conn); $num_rows = mysql_num_rows($result); $num_rows = $num_rows + 1; $id= $num_rows;

修改查询：

这是已经修改

$sql="INSERT INTO `webgeek1_software_order`.`application` (`dateAndTime`, `vehicleNo`, `description`,`amount`, `service`) VALUES('$dateAndTime', '$vehicleNo', '$description', '$amount', '$service')"; 

我相信数据库应该能够在插入新数据后自动设置下一个ID。这样可以避免因为您不再插入自己的号码而导致身份证复制等错误。

旁注（但很重要）：你应该使用mysqli或PDO statements因为mysql扩展已被弃用（和甚至在PHP 7.0.0删除）。

Answer 2

在phpmyadmin的结构为我的表，我需要设置列“的SerialNo”到AI（自动增量），并在该插入数据代码，我需要评论的行：

$num_rows = mysql_num_rows($result); 
$num_rows = $num_rows + 1; 
$id= $num_rows; 

类似地，我需要从插入查询中删除同一个文件中的'serialNo'。最后，我需要评论的线路：

echo "Serial No: " . " " . $id; 
echo "<br/>"; 

在插入数据代码