douyi6168 2015-05-25 10:37
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添加到mySQL数据库中的内容

I would like the current value in database in column mass_reach to be added with the new value under variable name $newmass (a constant figure). I created a double barrel loop below to extract what value is current then add it and update the value in second loop but cant seem to get this going.

if(isset($_POST['mass_reach'])){
   $massreach = $_POST['mass_reach'];

      foreach ($massreach as $massreach_id) {
        $query4  = "SELECT mass_reach FROM coaches WHERE id='{$massreach_id}'";       
        $result4 = mysqli_query($con, $query4);
        $query2  = "UPDATE coaches SET mass_reach='{$newmass}'+'{$result4}' WHERE id='{$massreach_id}'";          
        $result = mysqli_query($con, $query2);
        }
}

I am getting the error: I get the error: Catchable fatal error: Object of class mysqli_result could not be converted to string in line 8 ie

$query2 = "UPDATE coaches SET mass_reach='{$mass}'+'{$result4}' WHERE id='{$massreach_id}'";          
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  • du2229 2015-05-28 01:55
    关注
    function get_string_between($string, $start, $end){
     $string = " ".$string;
     $ini = strpos($string,$start);
     if ($ini == 0) return "";
     $ini += strlen($start);    
     $len = strpos($string,$end,$ini) - $ini;
     return substr($string,$ini,$len);}
    
     foreach ($mass_reach as $string) {
    
         $c_id = get_string_between($string,"[","]");
         $c_mass = get_string_between($string,"(",")");
         $query2  = "UPDATE coaches SET mass_reach='{$mass}' + '{$c_mass}' WHERE id='{$c_id}'";       
         $result2 = mysqli_query($con, $query2);
     }
    
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