Php 如何解决重复条目问题，并将一个条目插入mysql表？

那么您应该只插入一个，而不是2个

这就是为什么会有重复条目的原因

<?php
//open connection to mysql db
$connect = mysqli_connect("localhost","root"," ","tutorial") or die("Error " . mysqli_error($connect));


$offence_place = $_POST["offence_place"];
$vehicle_no = $_POST["vehicle_no"];
$offence_type = $_POST["offence_type"];
$offence_lotnumber = $_POST["offence_lotnumber"];
$offence_charges = $_POST["offence_charges"];



$query = " Insert into eSummon(offence_place, vehicle_no, offence_type, offence_lotnumber, offence_charges, image_name, image_path)
         values ('$offence_place','$vehicle_no','$offence_type','$offence_lotnumber','$offence_charges','$image_name','$path');";



mysqli_query($connect,$query) or die (mysqli_error($connect));

// IMAGE
header('Content-type : bitmap; charset=utf-8');


// Image Connection to Database
if (isset($_POST["encoded_string"])){

    $encoded_string = $_POST["encoded_string"]; 
    $image_name = $_POST["image_name"];         

    $decoded_string = base64_decode($encoded_string);   

    // Save image on the server
    $path = 'images/'.$image_name;

    $file = fopen($path, 'wb');                 

    $is_written = fwrite($file, $decoded_string);
    fclose($file);

    // Save the path to the Database
    if($is_written > 0) {

        // Open connection to mysql Database
        $connect = mysqli_connect("localhost","root"," ","tutorial") or die("Error " . mysqli_error($connect));

        $query = " Insert into eSummon(offence_place, vehicle_no, offence_type, offence_lotnumber, offence_charges, image_name, image_path)
        values ('$offence_place','$vehicle_no','$offence_type','$offence_lotnumber','$offence_charges','$image_name','$path');";


        $result = mysqli_query($connect, $query) or die("Error in Selecting " . mysqli_error($connect));

        if($result){
            echo "Success";
        }else{
            echo "Failed";
        }

        mysqli_close($connect);
    }


}



?>

删除一些顶级代码，并将代码转换为：

mysqli_query($connect,$query) --> you use this twice

---

通常，您会在重复密钥更新时使用
插入表
语法-假设存在某种类型的主键

<?php
if (isset($_POST["encoded_string"])){
$offence_place = $_POST["offence_place"];
$vehicle_no = $_POST["vehicle_no"];
$offence_type = $_POST["offence_type"];
$offence_lotnumber = $_POST["offence_lotnumber"];
$offence_charges = $_POST["offence_charges"];
$encoded_string = $_POST["encoded_string"]; 
$image_name = $_POST["image_name"];         

$decoded_string = base64_decode($encoded_string);   

// Save image on the server
$path = 'images/'.$image_name;

$file = fopen($path, 'wb');                 

$is_written = fwrite($file, $decoded_string);
fclose($file);

// Save the path to the Database
if($is_written > 0) {

    // Open connection to mysql Database
    $connect = mysqli_connect("localhost","root"," ","tutorial") or die("Error " . mysqli_error($connect));

    $query = " Insert into eSummon(offence_place, vehicle_no, offence_type, offence_lotnumber, offence_charges, image_name, image_path)
    values ('$offence_place','$vehicle_no','$offence_type','$offence_lotnumber','$offence_charges','$image_name','$path');";


    $result = mysqli_query($connect, $query) or die("Error in Selecting " . mysqli_error($connect));

    if($result){
        echo "Success";
    }else{
        echo "Failed";
    }

    mysqli_close($connect);
}


}   

?>

$sql=“插入'eSummon'（'officity\u place'、'vehicle\u no'、'officity\u type'、'officity\u lotnumber'、'officity\u charges'、'image\u name'、'image\u path`）
值（“$officity\u place”、“$vehicle\u no”、“$officity\u type”、“$officity\u lotnumber”、“$officity\u charges”、“$image\u name”、“$path”）
关于重复密钥更新
`违例地点“=$违例地点”，
`车辆编号“=”$车辆编号“，
`犯罪类型“=”$犯罪类型“，
`违例事项编号“=”$违例事项编号“，
`违例收费“=”违例收费“，
`image\u name`='$image\u name'，
`图像路径“=”$path“；”；
因此，我尝试使用您建议的代码。但在mqsql表中只显示映像_名称和映像_路径。当我希望插入所有数据时，其他记录（违例地点、车辆编号、违例类型、违例批号、违例费用）不显示。好的，谢谢！在我开始尝试您建议的代码之前，请允许我问一下您的意思：/*确保这些*/$path=$image\u name=''有一个默认值；我应该为这两个变量中的每一个输入值吗？不，但在您的初始查询中，我看不到这两个变量的声明位置-因此您有可能出错。对不起，您能给我一个示例，说明您的默认值是什么吗？这样我就可以从您的角度理解了。正如所写的一样-一个空字符串应该足够了，因为它将在第二个查询中更新解决！好的，谢谢你的帮助。非常感谢。
$sql="insert into `eSummon`(`offence_place`, `vehicle_no`, `offence_type`, `offence_lotnumber`, `offence_charges`, `image_name`, `image_path`)
                        values ( '$offence_place', '$vehicle_no', '$offence_type', '$offence_lotnumber', '$offence_charges', '$image_name', '$path')
                        on duplicate key update
                            `offence_place`='$offence_place',
                            `vehicle_no`='$vehicle_no',
                            `offence_type`='$offence_type',
                            `offence_lotnumber`='$offence_lotnumber',
                            `offence_charges`='$offence_charges',
                            `image_name`='$image_name',
                            `image_path`='$path';";



<?php
    /* Create db connection object */
    $connect = new mysqli( 'localhost', 'root', ' ', 'tutorial' ) or die('Error: unable to connect to db ');

    /* Get the variables assigned */
    $offence_place = $_POST['offence_place'];
    $vehicle_no = $_POST['vehicle_no'];
    $offence_type = $_POST['offence_type'];
    $offence_lotnumber = $_POST['offence_lotnumber'];
    $offence_charges = $_POST['offence_charges'];

    /* Ensure there is a default value for these */
    $path = $image_name='';

    /* Create the sql statement */
    $sql="insert into `eSummon`( `offence_place`, `vehicle_no`, `offence_type`, `offence_lotnumber`, `offence_charges`, `image_name`, `image_path` )
            values ( ?, ?, ?, ?, ?, ?, ? )
            on duplicate key update
                `offence_place`=?,
                `vehicle_no`=?,
                `offence_type`=?,
                `offence_lotnumber`=?,
                `offence_charges`=?,
                `image_name`=?,
                `image_path`=?;";

    /* Use aprepared statement */
    $stmt=$connect->prepare( $sql );
    $stmt->bind_params( 'sssssss', $offence_place,$vehicle_no,$offence_type,$offence_lotnumber,$offence_charges,$image_name,$path );
    $stmt->execute();


    /* Why this header? If you echo text further it will break the image! */
    header('Content-type: bitmap; charset=utf-8');



    if( isset( $_POST['encoded_string'] ) ){

        $encoded_string = $_POST['encoded_string']; 
        $image_name = $_POST['image_name'];
        $decoded_string = base64_decode( $encoded_string );

        $path = 'images/'.$image_name;
        $file = fopen( $path, 'wb' );                 
        $is_written = fwrite( $file, $decoded_string );
        fclose( $file );

        if( $is_written > 0 ) {

            /* New values have been assigned to image_name and path, execute statement again */
            $res=$stmt->execute();

            echo $res ? 'Success' : 'Failed';/* this would break the image */
        }
    }
?>