Php 继续把我弄出去
为什么我的第二个if从未被执行?这句话似乎能让我摆脱困境。我试过了,但没有成功Php 继续把我弄出去,php,continue,Php,Continue,为什么我的第二个if从未被执行?这句话似乎能让我摆脱困境。我试过了,但没有成功 foreach($columns as $i=>$column) { // Check if column exists $sql = "SELECT '$column' FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_SCHEMA = '$database' AND TABLE_NAME
foreach($columns as $i=>$column)
{
// Check if column exists
$sql = "SELECT '$column' FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_SCHEMA = '$database' AND TABLE_NAME = '$strTable'";
if(mysqli_real_query($link, $sql))
{
echo 'Column '.$column.' was created! <br>';
continue;
}
$sql = "alter table '$strTable' add column '$column' varchar (30)";
if(mysqli_real_query($link, $sql))
{
echo 'Column '.$column.' was created! <br>';
}
$cols .= $column.',';
}
您没有测试查询是否找到任何行。mysqli_real_查询只要没有出错就成功,但这并不意味着查询匹配了任何内容。您需要获得查询的结果 此外,您只是检查表是否存在,而不是检查表中是否存在该列 继续跳过循环体的整个其余部分。当找不到列时,应该使用else执行第二个块 如果要使用结果,请使用mysqli_查询。否则,需要先调用mysqli_use_result和mysqli_store_result;看
continue意味着开始循环的下一次迭代,跳过正文的其余部分。只需删除continue。其目的是不添加已存在的列。警告:mysqli_num_rows预期参数1为mysqli_result,bool给定$sql=从信息_SCHEMA.COLUMNS中选择1,其中表_SCHEMA='$database',表_NAME='$strTable',列_NAME='$COLUMN';
foreach($columns as $column)
{
// Check if column exists
$sql = "SELECT 1 FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_SCHEMA = '$database' AND TABLE_NAME = '$strTable' AND COLUMN_NAME = '$column'";
$result = mysqli_query($link, $sql);
if(mysqli_num_rows($result) > 0)
{
echo 'Column '.$column.' already exists! <br>';
} else {
$sql = "alter table '$strTable' add column '$column' varchar (30)";
if(mysqli_real_query($link, $sql))
{
echo 'Column '.$column.' was created! <br>';
}
}
$cols .= $column.',';
}