我尝试先通过创建表名来创建表。但我想我创建的表名称并不能帮助我的代码做到这一点，并且它给出了以下错误:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
这是我的代码:
<html>
<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<title>next change</title>
<style type="text/css">
.auto-style1 {
margin-left: 0px;
}
.auto-style2 {
margin-left: 13px;
}
.auto-style3 {
margin-left: 6px;
}
.auto-style4 {
margin-left: 276px;
}
</style>
</head>
<body>
<form action="" method="POST">
Session Starting Year : 20
<select name="session" style="width: 48px" class="auto-style1">
<option value="14">14</option>
<option value="15">15</option>
</select>
Program:
<select class="auto-style2" name="course" style="width: 141px">
<option value="mca">MCA</option>
<option value="btcse">B.Tech. CSE</option>
<option value="btece">B. Tech. ECE</option>
<option value="msccs">MSc. Comp. Sc.</option>
</select>
Semester:
<select name="semester" class="auto-style3" style="width: 55px">
<option value="sem1">I</option>
<option value="sem2">II</option>
<option value="sem3">III</option>
<option value="sem4">IV</option>
<option value="sem5">V</option>
<option value="sem6">VI</option>
<option value="sem7">VII</option>
<option value="sem8">VIII</option>
</select>
<br/>
<br/>
<input name="create" type="submit" value="Create" class="auto-style4" style="width: 77px" />
</form>
<?php
$table="";
if($_POST['create']){
$yr=$_POST['session'];
$course=$_POST['course'];
$sem=$_POST['semester'];
$table=$yr.$course.$sem;
}
$table=mysql_real_escape_string($table);
echo $table."<br/";
$dbName = "test";
$conn = mysql_connect('localhost','root','','test');
if (!$conn) {
die('Could not connect to MySQL: ' . mysql_error());
}
mysql_select_db($dbName) or die(mysql_error());
//Upload the csv file
if (isset($_POST['submit'])) {
if (is_uploaded_file($_FILES['filename']['tmp_name'])) {
echo "<h1>" . "File ". $_FILES['filename']['name'] ." uploaded successfully." . "</h1>";
}
//get the first row fields
$fields = "";
$fieldsInsert = "";
if (($handle = fopen($_FILES['filename']['tmp_name'], "r")) !== FALSE) {
if(($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
$num = count($data);
$fieldsInsert .= '(';
for ($c=0; $c < $num; $c++) {
$fieldsInsert .=($c==0) ? '' : ', ';
$fieldsInsert .="`".$data[$c]."`";
$fields .="`".$data[$c]."` varchar(10) DEFAULT NULL,";
}
$fieldsInsert .= ')';
}
//drop table if exist
if(mysql_num_rows(mysql_query("SHOW TABLES LIKE '".$table."'"))>=1) {
mysql_query('DROP TABLE IF EXISTS '.$table) or die(mysql_error());
}
//create table
$sql = "CREATE TABLE `".$table."` (
".$fields."
PRIMARY KEY (`roll`)
) ";
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not create table: ' . mysql_error());
}
else {
while(($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
$num = count($data);
$fieldsInsertvalues="";
//get field values of each row
for ($c=0; $c < $num; $c++) {
$fieldsInsertvalues .=($c==0) ? '(' : ', ';
$fieldsInsertvalues .="'".$data[$c]."'";
}
$fieldsInsertvalues .= ')';
//insert the values to table
$sql = "INSERT INTO ".$table." ".$fieldsInsert." VALUES ".$fieldsInsertvalues;
mysql_query($sql,$conn);
}
echo 'Table Created';
}
fclose($handle);
}
print "Import Done.";
}else {
print "Upload new csv by browsing to file and clicking on Upload<br />\n";
print "<form enctype='multipart/form-data' action='next_change.php' method='post'>";
print "File name to import:<br />\n";
print "<input size='50' type='file' name='filename'><br />\n";
print "<input type='submit' name='submit' value='Upload'></form>";
}
?>
</body>
</html>
我做错了什么？
最佳答案
您的 $fieldsInsert 变量应包含在“(”和“)”内。请参阅 mysql insert http://dev.mysql.com/doc/refman/5.7/en/insert.html 的语法
关于php - 使用 MySQL 和 PHP 在 CSV 文件上传时动态创建表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40938647/