查询将图像名称插入数据库



图像名称上传到 databse 时,使用查询插入的正确方法是什么?这是一个不起作用的短代码。

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

当前结构是 update sql,但您正在使用插入命令。您的插入将是这样的:

    $query='INSERT INTO table_name (images1,images2,images3,images4) 
        values("'.$_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

这应该是一个 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'] . '"';

插入应为:

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

插入查询就像

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

您正在我们使用column_name=vlaue的地方编写更新查询 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'].'")';

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