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Member Avatar for Mohammed_44

//The php section of the code <?php function getJSONFromDB($sql){ $conn = mysqli_connect("localhost", "root", "","rent"); //echo $sql; $result = mysqli_query($conn, $sql)or die(mysqli_error()); $arr=array(); while($row = mysqli_fetch_assoc($result)) { $arr[]=$row; } return json_encode($arr); } $jsonData= getJSONFromDB("select Pic from Display"); $jsn=json_decode($jsonData); //for loop to retrieve rows from database for($i=0;$i<sizeof($jsn);$i++){ $a=$jsn[$i]->Pic; //here is the main …

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echo "<head></head><div style='float:left; padding:30px'><h1>$k['Name']</h1></br><img src='sponsor/".$k['Pic'].'.jpeg'."' height='300' width='300'><br>.</hr></div>"; The error occured after i wanted to echo $k['Name']; This is the error that it shows Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING)

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