删除 SQL 数组 PHP 中的单个项目

Question

这让我发疯：如果我在 SQL 数据库中有这样的条目：[1,2,3,4]

如何删除数组中的单个项目？

mysqli_query = ("DELETE FROM table WHERE id='$id'... (something to specify the item here)");??

Answer 1

这个应该可以，当然要根据需要调整。

$string = cell date to here;

$new_string = str_replace('replace', 'with', $string);

echo "{$string}<br>{$new_string}";

// commented until new strings visually compared.
#mysqli_query("update table set `items` = {$new_string} WHERE `id` = {$id}");

我更喜欢在变量周围使用大括号，这只是一种偏好，我通常不会在 php 中使用 html 但是对于单个断行，跳出并返回 [= 似乎毫无意义16=].

Answer 2

我的解决方法是无稽之谈，STR_REPLACE 是门票。这是完整的 like/dislike 代码：

//LIKE FIELD  
    if(isset($_POST['like'.$id])) { 
    if (!in_array("$id", $like_explode)){
    if (!in_array("$id", $dislike_explode)){
    mysqli_query($db, "UPDATE likes SET pid_like=CONCAT(pid_like,',$id') WHERE user_id='$user_id'");  
    mysqli_query($db, "UPDATE comments SET likes=(likes+1) WHERE id='$id'"); 
    }   
    else 
    { 
    $new_dislike_string = str_replace(",$id", '', $dislike_string);
    mysqli_query($db, "UPDATE likes SET pid_dislike='$new_dislike_string' WHERE user_id='$user_id'");
    mysqli_query($db, "UPDATE likes SET pid_like=CONCAT(pid_like,',$id') WHERE user_id='$user_id'");
    mysqli_query($db, "UPDATE comments SET likes=(likes+1) WHERE id='$id'"); 
    mysqli_query($db, "UPDATE comments SET dislikes=(dislikes-1) WHERE id='$id'");
    }
    }  
    } 
//DISLIKE FIELD         
    if(isset($_POST['dislike'.$id])) { 
    if (!in_array("$id", $dislike_explode)){
    if (!in_array("$id", $like_explode)){
    mysqli_query($db, "UPDATE likes SET pid_dislike=CONCAT(pid_dislike,',$id') WHERE user_id='$user_id'");  
    mysqli_query($db, "UPDATE comments SET likes=(dislikes+1) WHERE id='$id'"); 
    }   
    else 
    { 
    $new_like_string = str_replace(",$id", '', $like_string);
    mysqli_query($db, "UPDATE likes SET pid_like='$new_like_string' WHERE user_id='$user_id'");
    mysqli_query($db, "UPDATE likes SET pid_dislike=CONCAT(pid_dislike,',$id') WHERE user_id='$user_id'");
    mysqli_query($db, "UPDATE comments SET dislikes=(dislikes+1) WHERE id='$id'"); 
    mysqli_query($db, "UPDATE comments SET likes=(likes-1) WHERE id='$id'");
    }
    }  
    }    
//LIKE-DISLIKE FIELD END

非常感谢，克里斯！

P.S。唯一的牺牲是将 pid_like/dislike 字段的默认值开始为“0”，或者进行一些令人讨厌的修剪以摆脱第一个“，”

删除 SQL 数组 PHP 中的单个项目

Delete single item in SQL array PHP

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sql-delete