mysql_fetch_assoc（）期望参数1是资源，对象给出[重复]

问题描述：

This question already has an answer here:

mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

Im trying to store the results from my query into a variable but get this error

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, object given

$result = mysqli_query($con,"Select objectID from objects where objectName = '$_POST[relations]'");


if($result === FALSE) {
    die(mysqli_error()); // TODO: better error handling
}

$result = mysql_fetch_assoc($result);
echo $result;

Any other way of storing this variable?

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此问题已经存在这里有一个答案： p>

mysql_fetch_array（）/ mysql_fetch_assoc（）/ mysql_fetch_row（）/ mysql_num_rows etc ...期望参数1为资源或结果 32 answers span> li> ul> div>

我试图将查询结果存储到变量中但得到此错误 p>

警告：mysql_fetch_assoc（）期望参数1为资源，给定对象 p>

  $ result = mysqli_query（$ con，“从objectName ='$ _POST [relations]''的对象中选择objectID）; 
 
 
 nif（$ result =  == FALSE）{
 die（mysqli_error（））;  // TODO：更好的错误处理
} 
 
 $ result = mysql_fetch_assoc（$ result）; 
echo $ result; 
  code>  pre> 
 
 任何其他存储方式 这个变量？ p> 
  div>

答

You start with mysqli_*, but continue with mysql_*.

Change mysql_fetch_assoc to mysqli_fetch_assoc.

Documentation
You can view the documentation on mysqli_fetch_assoc here.

mysql_fetch_assoc（）期望参数1是资源，对象给出[重复]

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