Question

将图片名称上传到数据库时使用查询插入的正确方法是什么？这是一个无效的简短代码。

$query = 'INSERT INTO table_name
    SET images1="' . $_FILES['file1']['name'] . '",
        images2="' . $_FILES['file2']['name'] . '",
        images3="' . $_FILES['file3']['name'] . '",
        images4="' . $_FILES['file4']['name'] . '"'

Answer 1

当前的结构是更新sql，但你正在使用insert命令。你的插入将是这样的：

    $query='INSERT INTO table_name (images1,images2,images3,images4) 
        values("'.$_FILES['file1']['name'].'", "'.$_FILES['file2']['name'].'",
 "'.$_FILES['file3']['name'].'", "'.$_FILES['file4']['name'].'")';

如果您想要更新，请使用：

    $query='update table_name SET images1="'.$_FILES['file1']['name'].'",
     images2="'.$_FILES['file2']['name'].'", images3="'.$_FILES['file3']['name'].'",
 images4="'.$_FILES['file4']['name'].'"' // add where clause if any

Answer 2

这应该是UPDATE查询：

$query = 'INSERT INTO table_name
    SET images1="' . $_FILES['file1']['name'] . '",
        images2="' . $_FILES['file2']['name'] . '",
        images3="' . $_FILES['file3']['name'] . '",
        images4="' . $_FILES['file4']['name'] . '"';

INSERT应该是：

$query = 'INSERT INTO table_name (images1, images2, images3, images4) 
    values( "' . $_FILES['file1']['name'] . '",
            "' . $_FILES['file2']['name'] . '",
            "' . $_FILES['file3']['name'] . '",
            "' . $_FILES['file4']['name'] . '"
)';

Answer 3

插入查询就像

INSERT INTO table_name (column1,column2,column3,...) VALUES (value1,value2,value3,...);

您正在使用column_name=vlaue编写更新查询所以你的查询将是

$query='INSERT INTO table_name (images1,images2,images3,images4) 
values("'.$_FILES['file1']['name'].'", "'.$_FILES['file2']['name'].'", "'.$_FILES['file3']['name'].'", "'.$_FILES['file4']['name'].'")';

查询将图像名称插入数据库

3 个答案: