PDO json插入多个标签 [英] PDO json insert multiple tags

问题描述

我一直在尝试使用pdo将一些标签插入表中，但无济于事.

I've been trying to insert some tags into a table using pdo but to no avail.

我有一个名为Tag的php数组.

I have a php array called Tag.

标签数组中的样本数据如下

Sample data in tag array is as follows

   tag[] = [[a,b,c,d,e],[f,g,h,i,j]]

使用下面的for循环将其转换为(1，'a'，'b'，'c'，'e'，0)，(1，'f'，'g'，' h'，'i'，0)

using a for loop below I'm able to convert it to (1,'a','b','c','e',0), (1,'f','g','h','i',0)

      $value="";
      $value .= "($postid,";
      for($i=0;$i<sizeof($tag);$i++)
      {
        $value .="'$tag[$i]'";
          if($i + 1 == $sizeof($tag){
          $value .=")";
         }else{
          $value .="),";
         }
      }

并按如下所示准备并插入表格中

And prepare and insert into the table as follows

      $inserttagquery = "insert Into tagtable  ( postid, desc, b, u, toppos,leftpos ,ver) values :value";
      $queryinserttag = $conn->prepare($inserttagquery);
      $queryinserttag->execute(array('value'=>$value));
      $insertedtag = $queryinserttag->rowCount();

但是，这似乎不起作用. $ insertedtag不返回任何值.

However, this does not seem to work. $insertedtag does not return any value.

推荐答案

您的SQL完全错误.您在字段列表中添加了6个字段 AND ，然后仅提供一个占位符来为这些字段提供值.

Your SQL is outright wrong. You're lising 6 fields AND a constant value in your field list, then providing only a SINGLE placeholder to provide values for those fields.

1. 您不能使用数字作为字段名称. 0是完全语法错误和无效的字段名称.
2. 占位符在字段和值之间具有1:1关系.您不能将多个值推送到一个变量中，并尝试将其与占位符一起使用以填写其他字段.

1. you cannot use numbers as a field name. 0 is a flat out syntax error and an invalid field name.
2. Placeholders have a 1:1 relation between a field and a value. You CANNOT shove multiple values into a single variable and try to use that value with a placeholder to fill in OTHER fields.

您的查询应该是:

INSERT INTO tagtable (postid, desc, b, u, etc...) 
VALUES (:postid, :desc, :b, :u, etc...)

，然后为每个保镖提供INDIVIDUAL值:

and then you provide INDIVIDUAL values for each placehodler:

 $stmt->execute(array($postid, $desc, $b, $u, etc...));

按照所写，并忽略所有其他问题，您的查询将尝试将(1, 'f', 'g', etc..)字符串推入postid字段中.

As written, and ignoring all the other problems, your query would try shove your (1, 'f', 'g', etc..) string into JUST the postid field.

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